Question

Let $C_1$ be the circle which passes through the origin and makes an intercept of $10$ on both the axes in the first quadrant. Let $C_2$ be the circle that touches both the coordinate axes and lies inside $C_1.$ Then, the maximum radius of $C_2$ is

• A. $10(2-\sqrt{2})$
• B. $10(2+\sqrt{2})$
• C. $10\sqrt{2}$
• D. $10(\sqrt{2}-1)$

Answer 1

The correct option is A $10(2-\sqrt{2})$


Equation of the circle $C_1$ is ${(x-5)}^2+{(y-5)}^2=50$
As $C_2$ touches both the axes, its centre lies on the line $x=y.$
Let $C_2$ touches the $x$-axis and $y$-axis at $(0,h)\&(h,0)$, respectively.

So, the equation of the circle $C_2$ is ${(x-h)}^2+{(y-h)}^2=h^2$
For maximum radius, $C_2$ touches $C_1$ internally.

$C_1{C}_2=|r_2-r_1|\Rightarrow \sqrt{(h-5{)}^2+(h-5{)}^2}=5\sqrt{2}-h\Rightarrow (h-5{)}^2+(h-5{)}^2=h^2-10\sqrt{2}h+50\Rightarrow h^2-(20-10\sqrt{2})h=0\Rightarrow h=10(2-\sqrt{2})$