The equation of the circle which cuts the circles $x^2+y^2+4 x-7=0$, $2 x^2+2 y^2+3 x+5 y-9=0, x^2+y^2+y=0$ orthogonally is

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Aicharya · Accepted Answer

The centre of the circle which cuts the given circles having equation $ \begin{aligned} & S_1: x^2+y^2+4 y-7=0 \\ & S_2: 2 x^2+2 y^2+3 x+5 y-9=0 \end{aligned} $ and $S_3: x^2+y^2+y=0$ is radical centre of circles $S_1, S_2$ and $S_3$. $\because$ Equation of radical axis of circles $S_1$ and $S_2$ is

So, coordinate of radical centre is $(2,1)$ and radius of required circle is equals to length of tangent drawn from radical centre $(2,1)$ to any one of the circle $ =\sqrt{4+1+8-7}=\sqrt{6} $ So, equation of required circle is : $ \begin{aligned} (x-2)^2+(y-1)^2 & =6 \\ \Rightarrow \quad x^2+y^2-4 x-2 y-1 & =0 . \end{aligned} $ Hence, options (a) is correct

The equation of the circle which cuts the circles $x^2+y^2+4 x-7=0$, $2 x^2+2 y^2+3 x+5 y-9=0, x^2+y^2+y=0$…

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