Chapter 12 Coordinate Geometry (Concepts)

Straight line

Coordinate geometry is a branch of geometry where the position of points is described using coordinates, and geometric figures are studied using algebraic equations. The straight line is one of the most basic and fundamental geometric objects in the Cartesian plane. In coordinate geometry, a straight line is defined as the locus of a point that moves in a constant direction, meaning its slope is constant. It can be represented by a linear equation involving the coordinates $x$ and $y$.

Before delving into the equations and properties of a straight line, let's recap some fundamental concepts and formulas from coordinate geometry that are essential tools for working with lines.

---

Basic Formulas in Coordinate Geometry

1. Distance Formula:

The Distance Formula is used to calculate the distance between any two points in a two-dimensional Cartesian coordinate system.

Let $P$ be a point with coordinates $(x_1, y_1)$ and $Q$ be another point with coordinates $(x_2, y_2)$. The distance between points $P$ and $Q$, denoted as $d(P, Q)$ or simply $d$, is given by:

Derivation:

Consider the points $P(x_1, y_1)$ and $Q(x_2, y_2)$ in the Cartesian plane. Draw a line segment connecting P and Q. Now, draw a horizontal line through P and a vertical line through Q. These two lines intersect at a point, let's call it R. The coordinates of R will be $(x_2, y_1)$.

The points P, Q, and R form a right-angled triangle PQR, with the right angle at R.

The length of the horizontal side PR is the absolute difference in the x-coordinates: $\text{PR} = |x_2 - x_1|$.

The length of the vertical side QR is the absolute difference in the y-coordinates: $\text{QR} = |y_2 - y_1|$.

The distance $d(P, Q)$ is the length of the hypotenuse PQ. By the Pythagorean theorem in the right-angled triangle PQR:

Taking the square root of both sides (and considering only the positive root as distance is non-negative):

This is the distance formula.

Answer:

---

2. Section Formula:

The Section Formula is used to find the coordinates of a point that divides a line segment joining two given points in a specific ratio. The division can be internal or external.

Let the line segment join points $P(x_1, y_1)$ and $Q(x_2, y_2)$. Let $R(x, y)$ be a point that divides the segment PQ in the ratio $m_1 : m_2$.

a) Internal Division:

If the point $R(x, y)$ lies on the line segment PQ (between P and Q), it divides PQ internally in the ratio $m_1 : m_2$. The coordinates of R are given by:

Derivation for Internal Division:

Let P, Q, and R be the points $(x_1, y_1)$, $(x_2, y_2)$, and $(x, y)$ respectively, such that R divides PQ internally in the ratio $m_1 : m_2$. This means $\frac{PR}{RQ} = \frac{m_1}{m_2}$.

Draw perpendiculars from P, R, and Q to the x-axis, meeting it at A, B, and C respectively. The coordinates of A, B, C are $(x_1, 0)$, $(x, 0)$, and $(x_2, 0)$.

Draw a line through P parallel to the x-axis, intersecting RB at D and QC at E.

Then $\text{PD} = \text{AB} = x - x_1$, $\text{RE} = \text{BC} = x_2 - x$.

Also, $\text{RD} = y - y_1$, $\text{QE} = y_2 - y$.

In $\triangle \text{PDR}$ and $\triangle \text{REQ}$, we have:

• $\angle \text{PDR} = \angle \text{REQ} = 90^\circ$ (Construction)
• $\angle \text{DRP} = \angle \text{ERQ}$ (Vertically opposite angles)
• $\angle \text{RPD} = \angle \text{QRE}$ (Corresponding angles, PD || RE || x-axis) - Correction: These are corresponding angles as PD and RE are parallel and PRQ is a transversal.

Thus, $\triangle \text{PDR} \sim \triangle \text{REQ}$ by AAA similarity.

Since the triangles are similar, the ratio of corresponding sides is equal:

We are given $\frac{PR}{RQ} = \frac{m_1}{m_2}$. So,

Cross-multiplying:

Similarly, using the ratio of the vertical sides:

Cross-multiplying:

Combining the x and y coordinates, we get the internal section formula (ii).

b) External Division:

If the point $R(x, y)$ lies on the line containing segment PQ but outside the segment (either on the extension of PQ or QP), it divides PQ externally in the ratio $m_1 : m_2$. The coordinates of R are given by:

Note: For external division, the ratio $m_1 : m_2$ implies that R is $m_1$ units away from P and $m_2$ units away from Q. If $m_1 > m_2$, R lies on the extension of PQ beyond Q. If $m_1 < m_2$, R lies on the extension of QP beyond P. The derivation is similar to internal division using similar triangles, but with careful attention to the signs of the ratios and segments. Alternatively, external division in ratio $m_1:m_2$ is equivalent to internal division in ratio $m_1:(-m_2)$. Substituting $-m_2$ for $m_2$ in the internal division formula gives the external division formula.

c) Midpoint Formula:

The midpoint of a line segment is a special case of internal division where the ratio is $1 : 1$ (i.e., $m_1 = 1$ and $m_2 = 1$). Substituting $m_1=1$ and $m_2=1$ into the internal section formula (ii):

This is the midpoint formula, which gives the coordinates of the midpoint of the segment joining $(x_1, y_1)$ and $(x_2, y_2)$.

---

3. Area of a Triangle:

The Area of a Triangle whose vertices are given by their coordinates can be calculated using a specific formula derived from coordinate geometry principles.

Let the vertices of a triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. The area of triangle ABC is given by:

The absolute value is used because the area of a geometric figure is always non-negative. The expression inside the absolute value can be positive or negative depending on the order in which the vertices are taken (clockwise or counterclockwise).

Alternative representation using Determinants:

This confirms the formula (v).

Condition for Collinearity of Three Points:

If three points $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ are collinear (lie on the same straight line), the area of the triangle formed by them is zero.

This condition is useful for checking if three points are collinear.

---

Slope of a Line

The Slope (or gradient) of a straight line is a numerical measure that describes the steepness and direction of the line. It is the ratio of the vertical change (rise) to the horizontal change (run) between any two distinct points on the line. The slope is constant for any given straight line.

1. Slope from Two Points:

If a line passes through two distinct points $P(x_1, y_1)$ and $Q(x_2, y_2)$, the slope of the line, denoted by $m$, is given by:

This formula is valid only if $x_1 \neq x_2$.

• If $x_1 = x_2$, the line is a vertical line parallel to the y-axis. The denominator is zero, so the slope is undefined.
• If $y_1 = y_2$ (and $x_1 \neq x_2$), the line is a horizontal line parallel to the x-axis. The numerator is zero, so the slope is $m = 0$.

Derivation:

Consider the points $P(x_1, y_1)$ and $Q(x_2, y_2)$. The change in y is $y_2 - y_1$, and the change in x is $x_2 - x_1$. The slope is defined as the ratio of these changes.

Draw a horizontal line through P and a vertical line through Q, meeting at R $(x_2, y_1)$. The slope is the ratio of the length QR to the length PR.

This simple ratio is the definition of the slope.

2. Slope from Angle of Inclination:

The angle of inclination of a straight line is the angle ($\theta$) formed by the line with the positive direction of the x-axis, measured counterclockwise. This angle is typically taken in the range $0^\circ \leq \theta < 180^\circ$.

The slope $m$ of a line is equal to the tangent of its angle of inclination.

This formula is valid provided $\theta \neq 90^\circ$ (since $\tan 90^\circ$ is undefined, corresponding to a vertical line).

Relationship between Slope and Angle of Inclination:

• If $0^\circ < \theta < 90^\circ$, $\tan \theta > 0$, so the slope $m$ is positive. The line goes upwards from left to right.
• If $\theta = 0^\circ$, $\tan \theta = 0$, so the slope $m$ is 0. The line is horizontal.
• If $90^\circ < \theta < 180^\circ$, $\tan \theta < 0$, so the slope $m$ is negative. The line goes downwards from left to right.
• If $\theta = 90^\circ$, $\tan \theta$ is undefined, the slope $m$ is undefined. The line is vertical.

Derivation:

Consider a line making an angle $\theta$ with the positive x-axis. Take any two distinct points $P(x_1, y_1)$ and $Q(x_2, y_2)$ on the line. From the definition of slope, $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Draw a line through P parallel to the x-axis. Draw a perpendicular from Q to this horizontal line, meeting it at R. The coordinates of R are $(x_2, y_1)$. The angle $\angle \text{QPR}$ is equal to the angle of inclination $\theta$ (corresponding angles).

In the right-angled triangle PQR, the side opposite to angle $\theta$ is $\text{QR} = y_2 - y_1$, and the side adjacent to angle $\theta$ is $\text{PR} = x_2 - x_1$.

The tangent of the angle $\theta$ in $\triangle \text{PQR}$ is given by:

Since $m = \frac{y_2 - y_1}{x_2 - x_1}$, we have $m = \tan \theta$.

---

Equations of a Straight Line

A straight line in the Cartesian plane can be represented by a linear equation in two variables $x$ and $y$. The form of the equation depends on the information given about the line (e.g., its slope, points it passes through, or its intercepts). Let's explore the different forms of the equation of a straight line.

1. Slope-Intercept Form:

This form is used when the slope ($m$) of the line and its y-intercept ($c$) are known. The y-intercept is the y-coordinate of the point where the line crosses the y-axis. This point has coordinates $(0, c)$.

The equation of a line with slope $m$ and y-intercept $c$ is given by:

Derivation:

Consider a line with slope $m$ and y-intercept $c$. This line passes through the point $(0, c)$.

Let $(x, y)$ be any arbitrary point on this line, distinct from $(0, c)$.

Using the formula for the slope of a line passing through two points (Formula (vi)) with $(x_1, y_1) = (0, c)$ and $(x_2, y_2) = (x, y)$:

Multiplying both sides by $x$:

Rearranging the terms, we get:

This is the equation of the line in slope-intercept form.

2. Point-Slope Form:

This form is used when the slope ($m$) of the line and the coordinates of a point $(x_1, y_1)$ that the line passes through are known.

The equation of a line with slope $m$ passing through the point $(x_1, y_1)$ is:

Derivation:

Consider a line with slope $m$ passing through the point $(x_1, y_1)$.

Let $(x, y)$ be any other arbitrary point on this line.

Since $(x_1, y_1)$ and $(x, y)$ are two points on the line, the slope of the line must be equal to the slope calculated using these two points. Using the slope formula (vi):

Assuming $x \neq x_1$, multiply both sides by $(x - x_1)$:

Rearranging the terms, we get the point-slope form:

If $x = x_1$ for all points on the line, then the line is vertical, its slope is undefined, and its equation is $x = x_1$. In this case, the point-slope form is not applicable.

3. Two-Point Form:

This form is used when the coordinates of two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ that the line passes through are known.

The equation of a line passing through two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ is:

This formula is valid provided $x_1 \neq x_2$ and $y_1 \neq y_2$.

• If $x_1 = x_2$, the line is vertical, and its equation is $x = x_1$.
• If $y_1 = y_2$, the line is horizontal, and its equation is $y = y_1$.

If $x_1 \neq x_2$, the slope of the line is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Derivation:

Consider the line passing through points $P(x_1, y_1)$ and $Q(x_2, y_2)$. Let $R(x, y)$ be any other arbitrary point on this line.

Since P, Q, and R lie on the same straight line, the slope of the segment PR must be equal to the slope of the segment PQ.

Slope of PR = $\frac{y - y_1}{x - x_1}$ (assuming $x \neq x_1$)

Slope of PQ = $\frac{y_2 - y_1}{x_2 - x_1}$ (assuming $x_1 \neq x_2$)

Equating the slopes:

This is the two-point form of the equation of a line. It is essentially the point-slope form where the slope $m$ is substituted by $\frac{y_2 - y_1}{x_2 - x_1}$ and $(x_1, y_1)$ is used as the known point.

4. Intercept Form:

This form is used when the x-intercept and the y-intercept of the line are known. The x-intercept is the x-coordinate of the point where the line crosses the x-axis (i.e., a point $(a, 0)$). The y-intercept is the y-coordinate of the point where the line crosses the y-axis (i.e., a point $(0, b)$).

The equation of a line with x-intercept $a$ (where $a \neq 0$) and y-intercept $b$ (where $b \neq 0$) is:

This form is not applicable if the line passes through the origin $(0, 0)$ because in that case, both intercepts are 0, and the equation $\frac{x}{0} + \frac{y}{0} = 1$ is undefined. Lines passing through the origin have the form $y=mx$ or $x=0$ or $y=0$.

Derivation:

Consider a line that makes x-intercept $a$ and y-intercept $b$. This means the line passes through the points $(a, 0)$ and $(0, b)$.

We can use the two-point form (Formula (x)) with $(x_1, y_1) = (a, 0)$ and $(x_2, y_2) = (0, b)$.

Cross-multiplying:

Rearranging the terms to get a positive constant term $ab$:

Now, divide the entire equation by $ab$ (since $a \neq 0, b \neq 0$, $ab \neq 0$):

This is the equation of the line in intercept form.

5. General Form of a Linear Equation:

Any straight line in the Cartesian plane can be represented by a linear equation of the form:

Where A, B, and C are real numbers, and A and B are not both zero ($A^2 + B^2 \neq 0$).

This is the most general form. All other forms can be converted to this form, and this form can be used to determine properties like slope and intercepts (if they exist).

Finding Slope and Intercepts from General Form:

Given the equation $Ax + By + C = 0$:

• If B $\neq$ 0: We can rearrange the equation to the slope-intercept form ($y = mx + c$).

  $\text{By} = -Ax - C$

  $\text{y} = -\frac{A}{B}x - \frac{C}{B}$

  Comparing this to $y = mx + c$, the slope is $m = -\frac{A}{B}$ and the y-intercept is $c = -\frac{C}{B}$.

• If B $=$ 0: The equation becomes $Ax + C = 0$. Since $A$ and $B$ are not both zero, if $B=0$, then $A$ must be non-zero ($A \neq 0$).

  $\text{Ax} = -C$

  $\text{x} = -\frac{C}{A}$

  This is the equation of a vertical line. The slope is undefined. It has an x-intercept at $(-\frac{C}{A}, 0)$, but no y-intercept (unless it is the y-axis itself, i.e., $A=1, C=0 \implies x=0$).

• If A $=$ 0: The equation becomes $By + C = 0$. Since $A$ and $B$ are not both zero, if $A=0$, then $B$ must be non-zero ($B \neq 0$).

  $\text{By} = -C$

  $\text{y} = -\frac{C}{B}$

  This is the equation of a horizontal line. The slope is $m=0$. It has a y-intercept at $(0, -\frac{C}{B})$, but no x-intercept (unless it is the x-axis itself, i.e., $B=1, C=0 \implies y=0$).

---

Angle Between Two Lines

Given two non-vertical lines with slopes $m_1$ and $m_2$, they intersect at a point and form two angles: an acute angle and an obtuse angle (which sum to $180^\circ$).

Let $\theta_1$ and $\theta_2$ be the angles of inclination of the two lines, so $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$.

The angle ($\phi$) between the two lines is given by the difference in their angles of inclination, i.e., $\phi = \theta_2 - \theta_1$ or $\phi = \theta_1 - \theta_2$.

Using the tangent subtraction formula, $\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$:

Substituting $m_1 = \tan \theta_1$ and $m_2 = \tan \theta_2$:

This formula gives the tangent of one of the angles between the lines. The other angle is $180^\circ - \phi$. The tangent of the other angle is $\tan(180^\circ - \phi) = -\tan \phi$.

If we want to find the acute angle between the lines, we take the absolute value of the tangent:

This formula is valid provided $1 + m_1 m_2 \neq 0$, which means $m_1 m_2 \neq -1$.

Conditions for Parallel and Perpendicular Lines (using Slopes):

The angle formula leads directly to the conditions for lines to be parallel or perpendicular:

• Parallel Lines: Two distinct non-vertical lines are parallel if and only if their slopes are equal.

  $\text{Condition for Parallelism:} \quad m_1 = m_2$

  ... (xv)

  If $m_1 = m_2$, then $m_2 - m_1 = 0$. From formula (xiii), $\tan \phi = 0$. This means $\phi = 0^\circ$ or $\phi = 180^\circ$. An angle of $0^\circ$ means the lines are coincident or parallel. If the lines are distinct, they are parallel. Vertical lines ($x = \text{constant}$) are parallel to each other and have undefined slopes.

• Perpendicular Lines: Two non-vertical lines are perpendicular if and only if the product of their slopes is -1.

  $\text{Condition for Perpendicularity:} \quad m_1 m_2 = -1$

  ... (xvi)

  If $m_1 m_2 = -1$, then $1 + m_1 m_2 = 0$. The denominator in formula (xiii) becomes zero, making $\tan \phi$ undefined. This occurs when $\phi = 90^\circ$ or $\phi = 270^\circ$. Since the angle of inclination is usually $0^\circ \leq \theta < 180^\circ$, the angle between the lines is $90^\circ$, meaning they are perpendicular. A vertical line ($m$ undefined) and a horizontal line ($m=0$) are also perpendicular.

---

Distance of a Point from a Line

The perpendicular distance from a given point to a given straight line is the shortest distance between the point and the line.

Let $P(x_1, y_1)$ be the given point, and let the equation of the line be in the general form $Ax + By + C = 0$. The perpendicular distance $d$ from point $P$ to the line is given by the formula:

The expression $|Ax_1 + By_1 + C|$ in the numerator is the absolute value of the result obtained by substituting the coordinates of the point $(x_1, y_1)$ into the expression $Ax + By + C$. The denominator $\sqrt{A^2 + B^2}$ is related to the normalization factor of the normal vector to the line.

Note: The derivation of this formula is a bit more involved and often done using methods involving projection or finding the intersection of the perpendicular line from the point to the given line. For the scope here, it's important to know and apply the formula.

---

Distance Between Two Parallel Lines

The distance between two parallel lines is the perpendicular distance between any point on one line and the other line. This distance is constant everywhere along the parallel lines.

Consider two parallel lines with equations in the general form. Since they are parallel, their slopes are equal, which means the coefficients of $x$ and $y$ are proportional. We can write their equations as:

$Ax + By + C_1 = 0$

$Ax + By + C_2 = 0$

Note that A and B are the same in both equations (if they are proportional, you can multiply one equation by a constant to make them equal). Also, $C_1 \neq C_2$, otherwise, the lines would be identical.

To find the distance between these lines, take any point on the first line $Ax + By + C_1 = 0$. A simple point can be found by setting $x=0$ (if $B \neq 0$), which gives $By + C_1 = 0 \implies y = -C_1/B$. So, a point is $(0, -C_1/B)$. If $B=0$, then $A \neq 0$, and the line is $Ax + C_1 = 0$, giving $x=-C_1/A$. A point is $(-C_1/A, 0)$. Let's use the point $(x_1, y_1)$ where $Ax_1 + By_1 + C_1 = 0$.

Now, find the perpendicular distance from this point $(x_1, y_1)$ to the second line $Ax + By + C_2 = 0$ using the distance of a point from a line formula (xvii):

Since $(x_1, y_1)$ lies on the first line $Ax + By + C_1 = 0$, we know that $Ax_1 + By_1 = -C_1$.

Substitute this into the numerator:

This is the formula for the distance between two parallel lines. Note that the absolute value $|C_2 - C_1|$ is the same as $|C_1 - C_2|$.

Circle

A circle is a fundamental shape in geometry, defined by a specific property related to distance. In coordinate geometry, we describe this property using equations involving the coordinates of points.

Formally, a circle is the locus of a point in a plane that moves such that its distance from a fixed point in the plane is always constant.

• The fixed point is called the centre of the circle.
• The constant distance is called the radius of the circle.

Let the fixed point (centre) be $C(h, k)$ and the constant distance (radius) be $r$. If $P(x, y)$ is any point on the circle, then the distance between $P$ and $C$ must be equal to $r$. We can express this relationship using the distance formula.

---

Equation of a Circle

The equation of a circle is the algebraic representation of the locus of a point satisfying the definition of a circle. Different forms of the equation are useful depending on the information given about the circle.

1. Standard Form (Centre-Radius Form):

This is the most intuitive form of the circle's equation, directly derived from its definition.

Let the centre of the circle be $C(h, k)$ and its radius be $r$. Let $P(x, y)$ be any point on the circle. By the definition of a circle, the distance between the centre $C(h, k)$ and any point $P(x, y)$ on the circle is equal to the radius $r$.

Using the distance formula between $C(h, k)$ and $P(x, y)$:

Since this distance must be equal to the radius $r$:

To eliminate the square root, square both sides of the equation:

This is the standard form or centre-radius form of the equation of a circle. It explicitly shows the coordinates of the centre $(h, k)$ and the square of the radius $r^2$.

Special Case: Circle with Centre at the Origin

If the centre of the circle is at the origin $(0, 0)$, i.e., $h=0$ and $k=0$, the standard equation simplifies to:

This is the equation of a circle centered at the origin with radius $r$.

Answer:

---

2. General Form:

The equation of a circle can also be expressed in a more general algebraic form. By expanding the standard form $(x - h)^2 + (y - k)^2 = r^2$, we get:

This equation is in the form of a quadratic equation in $x$ and $y$. We can simplify this by introducing new constants. Let $-2h = 2g$, $-2k = 2f$, and $h^2 + k^2 - r^2 = c$. Substituting these into the equation gives the general form of the equation of a circle:

Here, $g$, $f$, and $c$ are real constants. For this equation to represent a circle, it must be possible to rewrite it in the standard form $(x - h)^2 + (y - k)^2 = r^2$ with a positive radius squared.

Finding Centre and Radius from General Form:

Given the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we can find the coordinates of the centre and the radius by rearranging the terms and completing the square for the $x$ and $y$ variables.

Group the $x$ terms and $y$ terms together and move the constant term to the right side:

Complete the square for the $x$ terms by adding $(g)^2$ to both sides, and for the $y$ terms by adding $(f)^2$ to both sides. Remember that completing the square for $z^2 + 2az$ involves adding $a^2$ to form $(z+a)^2$.

Rewrite the terms in parentheses as perfect squares:

Comparing this equation to the standard form $(x - h)^2 + (y - k)^2 = r^2$, we can identify the centre and the radius:

• The centre is $(h, k)$. Comparing $(x - h)^2$ with $(x + g)^2 = (x - (-g))^2$, we get $h = -g$. Comparing $(y - k)^2$ with $(y + f)^2 = (y - (-f))^2$, we get $k = -f$.

  Centre $(h, k) = (-g, -f)$

  $\text{Centre} = (-\frac{1}{2} \times \text{coefficient of x}, -\frac{1}{2} \times \text{coefficient of y})$

• The square of the radius is $r^2$. Comparing $r^2$ with $g^2 + f^2 - c$, we get $r^2 = g^2 + f^2 - c$.

  The radius is $r = \sqrt{g^2 + f^2 - c}$.

  $\text{Radius } r = \sqrt{\left(\frac{\text{coefficient of x}}{2}\right)^2 + \left(\frac{\text{coefficient of y}}{2}\right)^2 - \text{constant term}}$

Nature of the Circle based on $g^2 + f^2 - c$:

The nature of the graph represented by the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ depends on the value of $g^2 + f^2 - c$:

• If $g^2 + f^2 - c > 0$: $r^2$ is positive, so $r = \sqrt{g^2 + f^2 - c}$ is a real, positive number. The equation represents a real circle with centre $(-g, -f)$ and radius $r$.

• If $g^2 + f^2 - c = 0$: $r^2$ is zero, so $r = 0$. The equation $(x + g)^2 + (y + f)^2 = 0$ is only satisfied by the single point $(-g, -f)$. This is called a point circle.

• If $g^2 + f^2 - c < 0$: $r^2$ is negative. The equation represents an imaginary circle. There are no real points $(x, y)$ that satisfy this equation because the sum of two squares of real numbers $(x+g)^2 + (y+f)^2$ cannot be negative.

Answer:

---

Other Forms and Properties of the Circle Equation

Besides the standard and general forms, there are other useful forms of the circle equation derived from specific geometric conditions.

1. Equation of a Circle with Endpoints of a Diameter:

If the endpoints of a diameter of a circle are given as $A(x_1, y_1)$ and $B(x_2, y_2)$, the equation of the circle can be found using the property that any point $P(x, y)$ on the circumference (other than A or B) forms a right angle ($\angle APB = 90^\circ$) with the diameter.

The slope of AP is $m_{AP} = \frac{y - y_1}{x - x_1}$ (if $x \neq x_1$).

The slope of BP is $m_{BP} = \frac{y - y_2}{x - x_2}$ (if $x \neq x_2$).

Since AP is perpendicular to BP, the product of their slopes is -1 (for non-vertical/horizontal lines, using condition (xvi) from Straight Line section):

Multiplying both sides by $(x - x_1)(x - x_2)$:

This equation also holds if $P$ is one of the endpoints A or B. This is the equation of the circle with the given endpoints as the diameter.

2. Equation of a Circle Touching the Coordinate Axes:

If a circle touches one or both of the coordinate axes, its radius is related to the coordinates of its centre.

• Circle touching the x-axis: If a circle with centre $(h, k)$ and radius $r$ touches the x-axis, the perpendicular distance from the centre to the x-axis is equal to the radius. The distance from $(h, k)$ to the x-axis ($y=0$) is $|k|$. So, $r = |k|$. The equation becomes:

  $(x - h)^2 + (y - k)^2 = k^2$

  If the centre is $(h, r)$ (assuming $k=r>0$, i.e., in upper half plane), the equation is $(x - h)^2 + (y - r)^2 = r^2$.

• Circle touching the y-axis: If a circle with centre $(h, k)$ and radius $r$ touches the y-axis, the perpendicular distance from the centre to the y-axis is equal to the radius. The distance from $(h, k)$ to the y-axis ($x=0$) is $|h|$. So, $r = |h|$. The equation becomes:

  $(x - h)^2 + (y - k)^2 = h^2$

  If the centre is $(r, k)$ (assuming $h=r>0$, i.e., in right half plane), the equation is $(x - r)^2 + (y - k)^2 = r^2$.

• Circle touching both x and y axes: If a circle with centre $(h, k)$ and radius $r$ touches both axes, then the distances from the centre to both axes are equal to the radius. So, $r = |h|$ and $r = |k|$. This implies $|h| = |k| = r$. The centre's coordinates will be $(\pm r, \pm r)$ depending on the quadrant. For example, if the centre is in the first quadrant, the centre is $(r, r)$. The equation becomes:

  $(x - r)^2 + (y - r)^2 = r^2$

  ... (v)

  In general, it is $(x \pm r)^2 + (y \pm r)^2 = r^2$.

Parabola

The parabola is one of the important curves studied in coordinate geometry, belonging to the family of conic sections. A conic section is a curve obtained by intersecting a cone with a plane. Depending on the angle of the plane, we get different shapes like circles, ellipses, parabolas, and hyperbolas.

Geometrically, a parabola is defined as the locus of a point that moves in a plane such that its distance from a fixed point (called the focus) is equal to its distance from a fixed straight line (called the directrix). The constant ratio of these distances is called the eccentricity, and for a parabola, the eccentricity is always equal to 1 ($e=1$).

Let's define the key terms associated with a parabola:

• Focus (F): A fixed point not on the directrix.

• Directrix (d): A fixed straight line.

• Axis of the Parabola: The straight line passing through the focus and perpendicular to the directrix. The parabola is symmetric about its axis.

• Vertex (V): The point where the parabola intersects its axis. It is the midpoint of the perpendicular segment from the focus to the directrix, and it is the closest point on the parabola to the directrix and the focus.

• Latus Rectum: A line segment passing through the focus, perpendicular to the axis of the parabola, with its endpoints on the parabola.

---

Standard Equations of a Parabola with Vertex at Origin

The standard equations of a parabola are derived based on the definition using the distance formula, assuming the vertex is at the origin $(0, 0)$ and the axis of the parabola lies along one of the coordinate axes. Let the distance from the vertex to the focus (and also from the vertex to the directrix) be denoted by '$a$', where $a > 0$.

Derivation of the Standard Equation ($y^2 = 4ax$):

Consider the case where the vertex is at the origin $(0, 0)$ and the axis of the parabola is the positive x-axis.

Since the vertex is at $(0, 0)$ and the distance from the vertex to the focus is $a$, the focus $F$ will be at $(a, 0)$ on the positive x-axis.

Since the vertex is the midpoint of the perpendicular from the focus to the directrix, and the vertex is $(0, 0)$ and the focus is $(a, 0)$, the directrix must be a vertical line perpendicular to the x-axis at a distance $a$ to the left of the vertex. The equation of the directrix is $x = -a$, which can be written as $x + a = 0$.

Let $P(x, y)$ be any point on the parabola. According to the definition, the distance from $P$ to the focus $F(a, 0)$ is equal to the distance from $P$ to the directrix $x = -a$.

The distance from $P(x, y)$ to $F(a, 0)$ using the distance formula (from section on Straight Line):

The distance from $P(x, y)$ to the directrix $x + a = 0$ is the perpendicular distance, which is the distance along a horizontal line from $P$ to the point on the directrix with the same y-coordinate. The point on the directrix is $M(-a, y)$.

By the definition of the parabola, $\text{PF} = \text{PM}$.

Squaring both sides:

Expand both sides:

Subtract $x^2$ and $a^2$ from both sides:

Add $2ax$ to both sides:

This is the standard equation of a parabola that opens to the right.

The Four Standard Forms (Vertex at Origin, Axis along Coordinate Axis):

By placing the focus on different parts of the coordinate axes, we get four standard forms:

1. $y^2 = 4ax$ (Opens Rightward)

• Equation: $\text{y}^2 = 4ax$ ($a > 0$)
• Vertex: $V(0, 0)$
• Focus: $F(a, 0)$
• Equation of Directrix: $x = -a$ or $x + a = 0$
• Axis of the Parabola: The x-axis (equation $y = 0$)
• Length of Latus Rectum: $4a$
• Endpoints of Latus Rectum: Found by substituting $x=a$ into the equation: $y^2 = 4a(a) = 4a^2 \implies y = \pm 2a$. So the endpoints are $(a, 2a)$ and $(a, -2a)$.

2. $y^2 = -4ax$ (Opens Leftward)

Here, the focus is on the negative x-axis at $(-a, 0)$, and the directrix is the vertical line $x=a$.

• Equation: $\text{y}^2 = -4ax$ ($a > 0$)
• Vertex: $V(0, 0)$
• Focus: $F(-a, 0)$
• Equation of Directrix: $x = a$ or $x - a = 0$
• Axis of the Parabola: The x-axis (equation $y = 0$)
• Length of Latus Rectum: $4a$ (Length is always positive)
• Endpoints of Latus Rectum: Found by substituting $x=-a$ into the equation: $y^2 = -4a(-a) = 4a^2 \implies y = \pm 2a$. So the endpoints are $(-a, 2a)$ and $(-a, -2a)$.

3. $x^2 = 4ay$ (Opens Upward)

Here, the focus is on the positive y-axis at $(0, a)$, and the directrix is the horizontal line $y=-a$.

• Equation: $\text{x}^2 = 4ay$ ($a > 0$)
• Vertex: $V(0, 0)$
• Focus: $F(0, a)$
• Equation of Directrix: $y = -a$ or $y + a = 0$
• Axis of the Parabola: The y-axis (equation $x = 0$)
• Length of Latus Rectum: $4a$
• Endpoints of Latus Rectum: Found by substituting $y=a$ into the equation: $x^2 = 4a(a) = 4a^2 \implies x = \pm 2a$. So the endpoints are $(2a, a)$ and $(-2a, a)$.

4. $x^2 = -4ay$ (Opens Downward)

Here, the focus is on the negative y-axis at $(0, -a)$, and the directrix is the horizontal line $y=a$.

• Equation: $\text{x}^2 = -4ay$ ($a > 0$)
• Vertex: $V(0, 0)$
• Focus: $F(0, -a)$
• Equation of Directrix: $y = a$ or $y - a = 0$
• Axis of the Parabola: The y-axis (equation $x = 0$)
• Length of Latus Rectum: $4a$
• Endpoints of Latus Rectum: Found by substituting $y=-a$ into the equation: $x^2 = -4a(-a) = 4a^2 \implies x = \pm 2a$. So the endpoints are $(2a, -a)$ and $(-2a, -a)$.

---

Answer:

---

Answer:

---

Parabolas with Vertex not at the Origin

If the vertex of the parabola is shifted from the origin $(0, 0)$ to a new point $(h, k)$, the equations of the parabola change accordingly. The shape and orientation relative to its axis remain the same, but the entire graph is translated.

The transformation from a coordinate system with origin at $(0,0)$ to a new system with origin at $(h, k)$ is given by replacing $x$ with $(x-h)$ and $y$ with $(y-k)$. Applying this transformation to the standard equations:

• Opens Rightward: The equation is $(y - k)^2 = 4a(x - h)$.

  Vertex: $(h, k)$, Focus: $(h+a, k)$, Directrix: $x = h-a$, Axis: $y = k$.

• Opens Leftward: The equation is $(y - k)^2 = -4a(x - h)$.

  Vertex: $(h, k)$, Focus: $(h-a, k)$, Directrix: $x = h+a$, Axis: $y = k$.

• Opens Upward: The equation is $(x - h)^2 = 4a(y - k)$.

  Vertex: $(h, k)$, Focus: $(h, k+a)$, Directrix: $y = k-a$, Axis: $x = h$.

• Opens Downward: The equation is $(x - h)^2 = -4a(y - k)$.

  Vertex: $(h, k)$, Focus: $(h, k-a)$, Directrix: $y = k+a$, Axis: $x = h$.

These forms allow us to work with parabolas whose vertices are not necessarily at the origin, extending the application of the standard forms. Expanding these equations will result in the general form of a parabola, which is a second-degree equation where either $x^2$ or $y^2$ is present, but not both $x^2$ and $y^2$ with the same coefficient (as in a circle) or both present (as in ellipse/hyperbola).