Question

The equation of the circle through the points of intersection of the circles $x ^ { 2 } + y ^ { 2 } - 6 x + 2 y + 4 = 0$

$y = x$

• A. $7 x ^ { 2 } + 7 y ^ { 2 } + 10 x - 10 y - 12 = 0$
• B. $7 x ^ { 2 } + 7 y ^ { 2 } - 10 x - 10 y - 12 = 0$
• C. $7 x ^ { 2 } + 7 y ^ { 2 } - 10 x + 10 y - 12 = 0$
• D. $7 x ^ { 2 } + 7 y ^ { 2 } + 10 x + 10 y + 12 = 0$

Answer 1

Answer: (B)

$7 x ^ { 2 } + 7 y ^ { 2 } - 10 x - 10 y - 12 = 0$

Answer 2

Equation of any circle through the points of intersection of given circles is

$\left( x ^ { 2 } + y ^ { 2 } - 6 x + 2 y + 4 \right) + \lambda \left( x ^ { 2 } + y ^ { 2 } + 2 x - 4 y - 6 \right) = 0$

$\Rightarrow$ $x ^ { 2 } ( 1 + \lambda ) + y ^ { 2 } ( 1 + \lambda ) - 2 x ( 3 - \lambda ) + 2 y ( 1 - 2 \lambda ) + ( 4 - 6 \lambda ) = 0$or, $x ^ { 2 } + y ^ { 2 } - \frac { 2 x ( 3 - \lambda ) } { ( 1 + \lambda ) } + \frac { 2 y ( 1 - 2 \lambda ) } { ( 1 + \lambda ) } + \frac { ( 4 - 6 \lambda ) } { ( 1 + \lambda ) } = 0$ …..(i)

Its centre $= \left\{ \frac { 3 - \lambda } { 1 + \lambda } , \frac { 2 \lambda - 1 } { 1 + \lambda } \right\}$ lies on the line y = x .

Then $\frac { 2 \lambda - 1 } { 1 + \lambda } = \frac { 3 - \lambda } { 1 + \lambda }$

$\{ \because \quad \lambda \neq - 1 \}$

$\Rightarrow$ $3 \lambda = 4$ $\Rightarrow \lambda = \frac { 4 } { 3 }$

Substituting the value of $\lambda = \frac { 4 } { 3 }$ in (i),

we get the required equation as

$7 x ^ { 2 } + 7 y ^ { 2 } - 10 x - 10 y - 12 = 0$