Coordinate Geometry, Loci and Analytic Regions Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Coordinate Geometry, Loci and Analytic Regions short notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Distance and Midpoint Formula

Two formulas that start most coordinate questions

Distance between

(x_{1}, y_{1})

and

(x_{2}, y_{2})

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Midpoint

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Distance is for length. Midpoint is for bisecting a side or diagonal.

Slope and Line Equation

Line toolkit

Slope between two points:

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Point-slope form:

y - y_{1} = m (x - x_{1})

x-axis intersection: put

y = 0

.
y-axis intersection: put

x = 0

Which Tool Should You Use?

Coordinate formula chooser

Question clue	First tool
Length / radius / side	Distance formula
Midpoint / bisects / diagonal midpoint	Midpoint formula
Line cuts x-axis or y-axis	Line equation, then substitute $y = 0$ or $x = 0$
Parallelogram vertex	Vector addition or equal diagonal midpoints

Parallelogram Vertex Shortcut

Parallelogram coordinate rule

In parallelogram

A B C D

A + C = B + D

Therefore:

D = A + C - B

Coordinate Geometry, Loci and Analytic Regions: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Ability MCQ

Let $C$ be the circle $x^{2} + y^{2} = 25$ . Tangents to $C$ at the points $(3, 4)$ and $(4, 3)$ intersect at $P$ . A third tangent to $C$ at the point $(- 5, 0)$ intersects the first two tangents at $Q$ and $R$ respectively. The area of the triangle $P QR$ is

A.
\frac{150}{7}
B.
\frac{120}{7}
C.
25
D.
15

Correct Answer:

Step-by-Step Solution

Key idea: This is a multi-step coordinate geometry question synthesising circle tangents and triangle area. The trap is assuming the circle is the incircle of $△ P QR$ and using $A r e a = r s$ , or getting bogged down in the distance formula for the side lengths. The fastest method is to find the vertices and use a vertical base.

Step 1: Find the equations of the tangents.

The tangent to $x^{2} + y^{2} = 25$ at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} = 25$ .

Tangent at $(3, 4)$ : $3 x + 4 y = 25$ .

Tangent at $(4, 3)$ : $4 x + 3 y = 25$ .

Tangent at $(- 5, 0)$ : $- 5 x = 25 ⟹ x = - 5$ .

Step 2: Find the vertices of $△ P QR$ .

$P$ is the intersection of $3 x + 4 y = 25$ and $4 x + 3 y = 25$ . By symmetry, $x = y$ .

$7 x = 25 ⟹ x = 25/7$ . So $P = (25/7, 25/7)$ .

$Q$ is the intersection of $3 x + 4 y = 25$ and $x = - 5$ .

$3 (- 5) + 4 y = 25 ⟹ 4 y = 40 ⟹ y = 10$ . So $Q = (- 5, 10)$ .

$R$ is the intersection of $4 x + 3 y = 25$ and $x = - 5$ .

$4 (- 5) + 3 y = 25 ⟹ 3 y = 45 ⟹ y = 15$ . So $R = (- 5, 15)$ .

Step 3: Calculate the area using the vertical base $QR$ .

The segment $QR$ lies on the vertical line $x = - 5$ .

Base length $= ∣15 - 10∣ = 5$ .

The height of the triangle is the horizontal distance from $P$ to the line $x = - 5$ .

Height $= ∣25/7 - (- 5) ∣ = 25/7 + 35/7 = 60/7$ .

Area $= \frac{1}{2} \times base \times height = \frac{1}{2} \times 5 \times \frac{60}{7} = \frac{150}{7}$ .

Trap avoided: The circle $C$ is tangent to all three sides, but it lies OUTSIDE the triangle (it is an excircle, not the incircle). Using the incircle formula $A r e a = r s$ would yield the wrong result. The base/height method bypasses this trap entirely.

Answer: \frac{150}{7}

Question 2 · Quantitative Ability MCQ

Let $C_{1}$ be the circle $x^{2} + y^{2} = 18$ . From a variable point $P$ on the line $L : x + y = 6$ , tangents are drawn to $C_{1}$ . The chord of contact of these tangents always passes through a fixed point $F$ . Let $C_{2}$ be the circle with centre $F$ and radius $5$ . The line $L$ divides $C_{2}$ into two regions. The area of the region containing the origin is

A.
\frac{25\pi}{2}
B.
25\pi
C.
\frac{25\pi}{4}
D.
18\pi

Correct Answer:

Step-by-Step Solution

Key idea: This is a pole-and-polar locus question synthesised with circle area. The key insight is recognising that the line $L$ is tangent to $C_{1}$ , which means the fixed point $F$ (the pole of $L$ ) is exactly the point of tangency.

Step 1: Find the fixed point $F$ .

Let $P = (a, 6 - a)$ be a point on the line $x + y = 6$ .

The chord of contact from $P$ to the circle $x^{2} + y^{2} = 18$ is given by $T = 0$ :

$a x + (6 - a) y = 18$

Rearrange to group by the parameter $a$ :

$a (x - y) + 6 y - 18 = 0$

For this line to pass through a fixed point for all values of $a$ , the coefficients must be zero:

$x - y = 0 ⟹ x = y$

$6 y - 18 = 0 ⟹ y = 3$

So the fixed point is $F (3, 3)$ .

Step 2: Understand the geometric meaning.

Notice that the distance from the origin to the line $x + y = 6$ is $\frac{6}{1 ^{2} + 1 ^{2}} = 32 = 18$ .

This means the line $L$ is exactly tangent to the circle $C_{1}$ at the point $(3, 3)$ .

A known theorem states that if a point $P$ moves along a tangent line to a circle, its polar (the chord of contact) always passes through the point of tangency. Thus, $F (3, 3)$ is simply the point of tangency.

Step 3: Analyse the new circle $C_{2}$ and the line $L$ .

$C_{2}$ has centre $F (3, 3)$ and radius $5$ .

The line $L$ is $x + y = 6$ . Does $L$ pass through the centre $F$ ?

Substitute $(3, 3)$ into $L$ : $3 + 3 = 6$ . Yes!

Since $L$ passes through the centre of $C_{2}$ , it is a diameter of $C_{2}$ .

Therefore, $L$ divides $C_{2}$ into two equal semicircles.

Step 4: Calculate the area.

The area of a semicircle of radius $5$ is $\frac{1}{2} π r^{2} = \frac{25 π}{2}$ .

The origin $(0, 0)$ satisfies $0 + 0 < 6$ , so it lies in one of these semicircles. The area of that region is exactly half the circle.

Answer: \frac{25\pi}{2}

More short notes in this unit

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Coordinate Geometry, Loci and Analytic Regions Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Coordinate Geometry, Loci and Analytic Regions short notes for CAT: 41 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice

A question from this chapter

Question 1

Let C be the circle x^{2} + y^{2} = 25 . Tangents to C at the points (3, 4) and (4, 3) intersect at P . A third tangent to C at the point (- 5, 0) intersects the first two tangents at Q and R respectively. The area of the triangle P QR is

Question 2

Let C_{1} be the circle x^{2} + y^{2} = 18 . From a variable point P on the line L : x + y = 6, tangents are drawn to C_{1} . The chord of contact of these tangents always passes through a fixed point F . Let C_{2} be the circle with centre F and radius 5 . The line L divides C_{2} into two regions. The area of the region containing the origin is

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