Question

A circle C of radius $1$ touches both the axes. Another circle of radius greater than C touches both the axes as well as the circle C. Then the radius of the other circle is

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

Answer 1

The correct option is A $3+2\sqrt{2}$

Centre of small circle is at $(r,r)$. $r$ is a radius of small circle
and centre of large circle is at $(R,R)$ . $R$ is a radius of large circle .

$\therefore \sqrt{(R-r{)}^2+(R-r{)}^2}=R+r$
$2(R^2+r^2-2Rr)=R^2+r^2+2rR$

$R^2+r^2-6Rr=0$

Given $r=1$

$R^2-6R+1=0$
$R=\frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2}$

Given $R>r.$

So, $R=3+2\sqrt{2}$