JEE-Mains 2014 (21/30)

Geometry Level 3

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

\frac{\sqrt{3}}{\sqrt{2}}

\frac{1}{4}

\frac{\sqrt{3}}{2}

\frac{1}{2}

1 solution

Vignesh Rao
Nov 16, 2017

Let the two circles be $C_1$ and $C_2$

$C_1 = x^2 + y^2 -2x-2y+1=0$

Let $C_2$ be centered at $(0,g)$ . Since $C_2$ passes through the origin it is of the form:

$C_2 = x^2 + y^2 -2gy=0$

Hence the radius of $C_2 = g$

Since the two circles externally touch each other, Distance between centers = Sum of radii

$\sqrt{1+(1-g)^2} = 1 + g$

On solving

$g = \frac{1}{4}$

JEE-Mains 2014 (21/30)

1 solution

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