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

Coordinate Geometry, Loci and Analytic Regions notes for CAT: 48 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Coordinate Geometry, Loci and Analytic Regions

CAT QA Geometry

Coordinate Geometry, Loci and Analytic Regions

A 3-step journey from points and lines to equations, curves, and bounded areas.

chapter PYQs

①

📍 t1 — Coordinate Geometry of Lines and Polygons

You learn distance, midpoint, line equations, parallelogram coordinates, and coordinate-area methods.

3 own-course PYQs Selected topic Moderate-high utility

②

⭕ t2 — Analytic Circles and Loci

You move from straight-line geometry to circle equations and distance-based moving points.

1 own-course PYQ

③

🧩 t3 — Inequality Regions and Coordinate Areas

You learn how inequalities shade regions and how boundaries combine to create areas.

2 own-course PYQs

End goal: convert coordinate information into distances, equations, intersections, and areas without depending on a perfect diagram.

Topic Hero: Coordinate Geometry of Lines and Polygons

Geometry → Coordinate Geometry → t1

Coordinate Geometry of Lines and Polygons

One-line hook: coordinates let you solve geometry by calculation, not by eye.

What you'll learn here

Distance, midpoint, and slope as geometry tools.
Line equation and x-axis / y-axis intersection methods.
Parallelogram vertex shortcuts using vectors or diagonals.
Coordinate triangle side lengths and inradius.
Polygon area through coordinate formulas.

Coordinates Are Addresses of Points

A point is an address: $(x, y)$

↔️

$x$ coordinate

How far right or left from the y-axis.

↕️

$y$ coordinate

How far up or down from the x-axis.

Exam habit: never trust a rough diagram more than the given coordinates.

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.

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}

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

Chapter Roadmap: Coordinate Geometry, Loci and Analytic Regions

Coordinate Geometry, Loci and Analytic Regions

📍 t1 — Coordinate Geometry of Lines and Polygons

⭕ t2 — Analytic Circles and Loci

🧩 t3 — Inequality Regions and Coordinate Areas

Topic Hero: Coordinate Geometry of Lines and Polygons

Coordinate Geometry of Lines and Polygons

What you'll learn here

Coordinates Are Addresses of Points

A point is an address: $(x, y)$

$x$ coordinate

$y$ coordinate

Distance and Midpoint Formula

Two formulas that start most coordinate questions

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

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit

A question from this chapter

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Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

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Unlock the whole course

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

Chapter Roadmap: Coordinate Geometry, Loci and Analytic Regions

Coordinate Geometry, Loci and Analytic Regions

📍 t1 — Coordinate Geometry of Lines and Polygons

⭕ t2 — Analytic Circles and Loci

🧩 t3 — Inequality Regions and Coordinate Areas

Topic Hero: Coordinate Geometry of Lines and Polygons

Coordinate Geometry of Lines and Polygons

What you'll learn here

Coordinates Are Addresses of Points

A point is an address: (x,y)

x coordinate

y coordinate

Distance and Midpoint Formula

Two formulas that start most coordinate questions

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

Step-by-Step Solution

Step-by-Step Solution

More notes in this unit

A question from this chapter

Login to view the complete notes

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

A point is an address: $(x, y)$

$x$ coordinate

$y$ coordinate