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The equation of the circle through the points of intersection of the circles

$\mathrm{x^2+y^2-6 x+2 y+4=0, \quad x^2+y^2+2 x-4 y-6=0}$ and with its centre on the line $\mathrm{y=x}$
Option: 1
$\mathrm{7 x^2+7 y^2+10 x-10 y-12=0}$

Option: 2
$\mathrm{7 x^2+7 y^2-10 x-10 y-12=0}$

Option: 3
$\mathrm{7 x^2+7 y^2-10 x+10 y-12=0}$

Option: 4
$\mathrm{7 x^2+7 y^2+10 x+10 y+12=0}$

Answers (1)

best_answer

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

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

$\mathrm{x^2(1+\lambda)+y^2(1+\lambda)-2 x(3-\lambda)+2 y(1-2 \lambda)+(4-6 \lambda)=0}$

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

$\mathrm{\begin{gathered} \text { Its centre }=\left\{\frac{3-\lambda}{1+\lambda}, \frac{2 \lambda-1}{1+\lambda}\right\} \quad \text { lies on the line } y=x \text {. Then } \frac{2 \lambda-1}{1+\lambda}=\frac{3-\lambda}{1+\lambda} \\ \end{gathered}}$

$\mathrm{\Rightarrow \quad 2 \lambda-1=3-\lambda}$ $\mathrm{\{\quad \lambda \neq-1\}}$

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

Substituting the value of $\mathrm{\lambda=\frac{4}{3}}$ in (i), we get the required equation as $\mathrm{7 x^2+7 y^2-10 x-10 y-12=0}$

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