Question

The equation of the circle which passes through the points $( 3 , - 2 )$ and $( - 2,0 )$ and centre lies on the line $2 x - y = 3$, is.

• A. $x ^ { 2 } + y ^ { 2 } - 3 x - 12 y + 2 = 0$
• B. $x ^ { 2 } + y ^ { 2 } - 3 x + 12 y + 2 = 0$
• C. $x ^ { 2 } + y ^ { 2 } + 3 x + 12 y + 2 = 0$
• D. None of these

Answer 1

Answer: (C)

$x ^ { 2 } + y ^ { 2 } + 3 x + 12 y + 2 = 0$

Answer 2

Let centre be (h, k), then

$( h - 3 ) ^ { 2 } + ( k + 2 ) ^ { 2 } = ( h + 2 ) ^ { 2 } + k ^ { 2 }$ $\Rightarrow 10 h - 4 k - 9 = 0$

Also the centre lies on the given line, so $2 h - k = 3$ .

On solving $k = - 6 , h = - \frac { 3 } { 2 }$

Radius is $( h - 3 ) ^ { 2 } + ( k + 2 ) ^ { 2 } = \frac { 145 } { 4 }$ , which is true for option (3) only.