GATE Chemical 2026 | GA | Question: 10

Question

As shown in the figure, circle $C_{1}$ with center $O_{1}$ and radius $r_{1}$ touches the square $V W X Y$ at points $P$ and $Q$ while circle $C_{2}$ with center $O_{2}$ and radius $r_{2}$ touches the square $V W X Y$ at points $R$ and $S$. The two circles touch each other at $T$.

Given $r_{1}=1 \mathrm{~cm}$ and $\overline{V Y}=\overline{V W}=4 \mathrm{~cm}, r_{2}=$ $\_\_\_\_$ cm.

• A. $4-3 \sqrt{2}$
• B. $1+2 \sqrt{2}$
• C. $7-4 \sqrt{2}$
• D. $5+3 \sqrt{2}$

Answer 1

First All Evualte thier  corrdinate of centere for both the triangle   that is O1, O2 respectively.

r1 = 1cm (Given )

 side of square  = 4cm . (Given )

O1(1,1) and O2 (4-r2 , 4 - r2)

distance(O1,O2) = r1 + r2  = 1 + r2  (i)

By distance Formula

Distance(O1, O2) = Root Under( (x2 - x1)^2 + (y2- y1)^2 ) = Root((4-r2) - 1 ) ^2 + (4 - r2 ) - 1 ) ^2

Root(2(3 - r2)^2 )

3Root2 - 1 = (1 + root2 ) r2

r2 = 7 - 4root2  

So option C  is correct