The radius of the circle which cuts the circles $x^2+y^2-4 x$ $-4 y+7=0, x^2+y^2+4 x-4 y+6=0$ and $x^2+y^2+4 x+$ $4 y+5=0$ orthogonally is

Question

Aicharya · Accepted Answer

Let equation circle be $x^2+y^2+2 g x+2 f y+c=0$
It is orthogonal with $x^2+y^2-4 x-4 y+7=0$ $\begin{aligned} & \therefore 2 g(-2)+2 f(-2)=c+7 \\ & \Rightarrow 4 g+4 f+c=-7...(i) \end{aligned}$
Also orthogonal with $x^2+y^2+4 x-4 y+6=0$ $\begin{aligned} & \therefore 2 g(2)+2 f(-2)=c+6 \\ & \Rightarrow 4 g-4 f-c=6...(ii) \end{aligned}$
And orthogonal with $\begin{aligned} & x^2+y^2+4 x+4 y+5=0 \\ & \therefore 2 g(2)+2 f(2)=c+5 \\ & \Rightarrow 4 g+4 f-c=5...(iii) \end{aligned}$
Solving (i), (ii) and (iii) we get $\begin{aligned} & \mathrm{c}=-6, f=\frac{-1}{8}, g=\frac{-1}{8} \\ & \text { radius }=\sqrt{g^2+f^2-\mathrm{C}}=\sqrt{\frac{1}{64}+\frac{1}{64}+6}=\frac{\sqrt{193}}{4 \sqrt{2}} \end{aligned}$

The radius of the circle which cuts the circles $x^2+y^2-4 x$ $-4 y+7=0, x^2+y^2+4 x-4 y+6=0$ and $x^2+y^2+4…

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