Question

Let C be the circle with centre at (1, 1) and radius 1. If T is the circle centred at (0, y) passing through origin and touching the circle C externally, then the radius of T is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{\sqrt{3}}{\sqrt{2}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is B

$\frac{1}{4}$

Use the property, when two circles touch each other externally, then the distance between the centre is equal to the sum of their radii, to get the required radius.
Let the coordinate of the centre of T be (0, k).
Distance between their centres
$k+1=\sqrt{1+(k-1{)}^2}$ $[\because C_1{C}_2=k+1]$
$\Rightarrow k+1=\sqrt{1+k^2+1-2k}$

$\Rightarrow k+1=\sqrt{k^2+2-2k}\Rightarrow k^2+1+2k=k^2+2-2k\Rightarrow k=\frac{1}{4}$
So, the radius of circle T is k, i.e. $\frac{1}{4}$.