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GATE 2026 CY – Question 10

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

As shown in the figure, circle C1 with center O1 and radius r1 touches the square VWXY at points P and Q while circle C2 with center O2 and radius r2 touches the square VWXY at points R and S. The two circles touch each other at T. Given r1 = 1 cm and VY̅̅̅̅ = VW̅̅̅̅̅ = 4 cm, r2 = _____ cm.

Two tangent circles touching opposite corners of a 4 cm square.
  1. 4 −3√2
  2. 1 + 2√2
  3. 7 −4√2
  4. 5 + 3√2

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Show answer and explanation

Correct answer: (C) 7 −4√2

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 $r_1$ from each of those sides. Circle $C_2$ (radius $r_2$) is tangent to the two sides at the opposite corner.

**Distance of each centre from its corner.** A circle of radius $r$ tangent to both sides at a corner has its centre on the diagonal at a distance $r\sqrt2$ from the 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.)