Question

Two circles of radii '$a$' and '$b$' touching each other externally, are inscribed in the area bounded by $y=\sqrt{1-x^2}$ and the x-axis. If $b=\frac{1}{2}$, then a is equal to

• A. $\frac{1}{4}$
• B. $\frac{1}{8}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{1}{4}$

Given area $x^2+y^2=1$ and x-axis

i.e. upper semicircle

According to the diagram

$B_{1}D=B_{2}D=1$

$C_{1}D=1-a,C_{2}D=1-b=\frac{1}{2}$

$A_1{C}_1=a,A_2{C}_2=b=\frac{1}{2}$

$\angle C_1{A}_{1}D=\angle C_2{A}_{2}D={90}^{\circ }$

By Pythagoras theorem

$A_{1}D=\sqrt{(C_{1}D{)}^2–(A_1{C}_1{)}^2}$

$=\sqrt{(1–a{)}^2–a^2}=\sqrt{1-2a}$

$A_{2}D=\sqrt{\frac{1}{4}-\frac{1}{4}}=0$

$C_1=(A_{1}D,a)=(\sqrt{1–2a},a)$

$C_2=(A_{2}D,b)=(0,\frac{1}{2})$

Since the circles touch each other

$C_1{C}_2=r_1+r_2$

$C_1{C}_2^2=(a+\frac{1}{2}{)}^2$

$⟹(1–2a)+(a-\frac{1}{2}{)}^2=(a+\frac{1}{2}{)}^2$

$⟹1–2a=2a$

$⟹a=\frac{1}{4}$