GATE 2026 IN GA Q.10: solution and answer

GATE 2026 · INGA Q.10 · 2 marks · MCQ

Circles $C_{1}$ , $C_{2}$ , and $C_{3}$ , with centers $O_{1}$ , $O_{2}$ , and $O_{3}$ , and radii $r_{1}$ , $r_{2}$ , and $r_{3}$ , respectively, touch each other as shown in the following figure. Given $r_{1} = 2$ cm, $r_{2} = 1$ cm and the angle $∠ O_{1} O_{3} O_{2}$ is $9 0^{\circ}$ , $r_{3} =$ ______ cm.

GATE 2026 IN GA question 10: Three circles, each touching the other two externally. — Figure from the GATE 2026 IN paper. It is part of the question.

(A)
$\frac{1}{2} (- 3 + 17)$
(B)
$\frac{1}{2} (3 + 17)$
(C)
$\frac{1}{2} (- 2 + 17)$
(D)
$\frac{1}{2} (- 3 + 217)$

Answer and solution

ANSWER

$r_{3} = \frac{1}{2} (- 3 + 17) \approx 0.56 cm$ , option (A).

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

When two circles touch externally, the distance between their centres equals the sum of their radii. Let $r_{3} = r$ .
Apply this to each pair. $O_{1} O_{3} = r_{1} + r = 2 + r$ and $O_{2} O_{3} = r_{2} + r = 1 + r$ . The circles "touch each other", and the figure shows $C_{1}$ and $C_{2}$ touching too, so $O_{1} O_{2} = r_{1} + r_{2} = 2 + 1 = 3 cm$ .
Triangle $O_{1} O_{3} O_{2}$ has its right angle at $O_{3}$ , so the side opposite it, $O_{1} O_{2}$ , is the hypotenuse. By Pythagoras: $(2 + r)^{2} + (1 + r)^{2} = 3^{2}$
Expand: $4 + 4 r + r^{2} + 1 + 2 r + r^{2} = 9 \Rightarrow 2 r^{2} + 6 r - 4 = 0 \Rightarrow r^{2} + 3 r - 2 = 0$
By the quadratic formula: $r = \frac{- 3 \pm 3 ^{2} + 4 \cdot 2}{2} = \frac{- 3 \pm 17}{2}$
A radius must be positive. $\frac{- 3 - 17}{2}$ is negative, so take the + sign: $r_{3} = \frac{1}{2} (- 3 + 17) \approx 0.56 cm$ . This is option (A).
Check: with $r \approx 0.5616$ , $(2.5616)^{2} + (1.5616)^{2} \approx 6.5616 + 2.4384 = 9 = 3^{2}$ , so the angle at $O_{3}$ is a right angle. Also $r_{3} \approx 0.56 cm$ is smaller than both $r_{1}$ and $r_{2}$ , as drawn.

KEY IDEA

When circles touch externally, the distance between two centres is the sum of their radii. That gives all three sides of triangle $O_{1} O_{3} O_{2}$ in terms of $r_{3}$ : $2 + r_{3}$ , $1 + r_{3}$ and $3$ . The right angle at $O_{3}$ makes $O_{1} O_{2} = 3$ the hypotenuse, and Pythagoras then gives a quadratic in $r_{3}$ .

WATCH OUT FOR

The usual slip is to put the hypotenuse in the wrong place. The right angle is at $O_{3}$ , so the hypotenuse is $O_{1} O_{2} = 3$ , not one of the sides that involve $r_{3}$ . Other errors are forgetting that $C_{1}$ and $C_{2}$ also touch (so $O_{1} O_{2} = r_{1} + r_{2}$ ), and picking option (B), $\frac{1}{2} (3 + 17)$ , after a sign slip in the quadratic formula. Option (B) is about 3.56 cm, larger than both given radii. That contradicts both the figure and Pythagoras, since the legs $2 + r_{3}$ and $1 + r_{3}$ could not then fit a hypotenuse of 3.

From the GATE 2026 IN paper. Every question of this paper · Download the full paper · All 22 Geometry questions, with notes

GATE 2026 IN GA Q.10: solution and answer

Answer and solution

More Geometry questions