Question

The equation of the circle through the points of intersection of $x ^ { 2 } + y ^ { 2 } - 1 = 0 , x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 1 = 0$ and touching the line $x + 2 y = 0$ is

• A. $x ^ { 2 } + y ^ { 2 } + x + 2 y = 0$
• B. $x ^ { 2 } + y ^ { 2 } - x + 20 = 0$
• C. $x ^ { 2 } + y ^ { 2 } - x - 2 y = 0$
• D. $2 \left( x ^ { 2 } + y ^ { 2 } \right) - x - 2 y = 0$

Answer 1

Answer: (C)

$x ^ { 2 } + y ^ { 2 } - x - 2 y = 0$

Answer 2

Family of circles is

$x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 1 + \lambda \left( x ^ { 2 } + y ^ { 2 } - 1 \right) = 0$ $x ^ { 2 } + y ^ { 2 } - \frac { 2 } { 1 + \lambda } x - \frac { 4 } { 1 + \lambda } y + \frac { 1 - \lambda } { 1 + \lambda } = 0$

Centre is $\left[ \frac { 1 } { 1 + \lambda } , \frac { 2 } { 1 + \lambda } \right]$ and radius

$= \sqrt { \left( \frac { 1 } { 1 + \lambda } \right) ^ { 2 } + \left( \frac { 2 } { 1 + \lambda } \right) ^ { 2 } - \left( \frac { 1 - \lambda } { 1 + \lambda } \right) } = \sqrt { \frac { 4 + \lambda ^ { 2 } } { ( 1 + \lambda ) ^ { 2 } } }$

Since it touches the line $x + 2 y = 0$

Hence Radius = perpendicular distance from centre to the line $= \left| \frac { \frac { 1 } { 1 + \lambda } + \frac { 4 } { 1 + \lambda } } { \sqrt { 1 ^ { 2 } + 2 ^ { 2 } } } \right|$

$= \sqrt { \frac { 4 + \lambda ^ { 2 } } { ( 1 + \lambda ) ^ { 2 } } } = \frac { \sqrt { 4 + \lambda ^ { 2 } } } { 1 + \lambda }$

$\Rightarrow \quad \sqrt { 5 } = \sqrt { 4 + \lambda ^ { 2 } }$

$\Rightarrow \lambda = \pm 1$

$\lambda = 1$.

$\therefore$ Equation of circle is $x ^ { 2 } + y ^ { 2 } - x - 2 y = 0$