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Question

(b) The equation of required circle is ${S_1} + \lambda {S_2} = 0$.

${x^2}(1 + \lambda ) + {y^2}(1 + \lambda ) + x(2 + 13\lambda ) - y\left( {\frac

Vedclass · Accepted Answer

$4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$

Vedclass · Answer

(b) The equation of required circle is ${S_1} + \lambda {S_2} = 0$.

${x^2}(1 + \lambda ) + {y^2}(1 + \lambda ) + x(2 + 13\lambda ) - y\left( {\frac{7}{2} + 3\lambda } \right) - \frac{{25}}{2} = 0$

Centre = $\left( {\frac{{ - (2 + 13\lambda )}}{2},\,\,\frac{{\frac{7}{2} + 3\lambda }}{2}} \right)$

Centre lies on $13x + 30y = 0$

$ \Rightarrow $$ - 13\left( {\frac{{2 + 13\lambda }}{2}} \right) + 30\left( {\frac{{\frac{7}{2} + 3\lambda }}{2}} \right) = 0$

$ \Rightarrow \,$$\lambda = 1$.

Hence the equation of required circle is $4{x^2} + 4{y^2} + 30x - 13y - 25 = 0.$

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

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