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GATE 2026 PH – Question 10

General Aptitude · Quantitative Aptitude: Algebra, Geometry and Mensuration · 2 marks · Multiple choice

As shown in the figure, circle $C_1$ with center $O_1$ and radius $r_1$ touches the square VWXY at points P and Q while circle $C_2$ with center $O_2$ and radius $r_2$ touches the square VWXY at points R and S. The two circles touch each other at T. Given $r_1=1$ cm and $\overline{VY}=\overline{VW}=4$ cm, $r_2=$ _____ cm.

Two externally tangent circles touching opposite corners of a square of side 4 cm.
  1. $4-3\sqrt2$
  2. $1+2\sqrt2$
  3. $7-4\sqrt2$
  4. $5+3\sqrt2$

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Show answer and explanation

Correct answer: (C) $7-4\sqrt2$

Explanation

Place the square VWXY with side 4 cm. Circle $C_1$ (radius $r_1=1$) is tangent to two sides meeting at a corner, so its centre is at distance $r_1\sqrt2$ from that corner along the diagonal. Circle $C_2$ (radius $r_2$) is tangent to the two sides at the opposite corner, so its centre is at distance $r_2\sqrt2$ from that corner.

**The centres lie on the same diagonal**, whose length is $4\sqrt2$. The circles touch externally at T, so the distance between the centres is $r_1+r_2$:
$$r_1\sqrt2+(r_1+r_2)+r_2\sqrt2=4\sqrt2 .$$

With $r_1=1$:
$$\sqrt2+1+r_2+\sqrt2\,r_2=4\sqrt2\;\Rightarrow\;r_2(1+\sqrt2)=3\sqrt2-1 .$$

$$r_2=\frac{3\sqrt2-1}{\sqrt2+1}=(3\sqrt2-1)(\sqrt2-1)=6-3\sqrt2-\sqrt2+1=\mathbf{7-4\sqrt2}\ \text{cm}.$$

(Numerically $7-5.657=1.343$ cm.)