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Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^2+y^2=36$ is one of those circles, then the other circle is
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Verified Answer
The correct answer is:
$x^2+y^2-18 x+36=0$
We have, equation of one circle
$$
\begin{aligned}
& x^2+y^2=36 \\
& \Rightarrow \quad x^2+y^2-36=0 \\
&
\end{aligned}
$$
and radical axis of two circle is $x-4=0$
So, equation of other circle is
$$
\begin{aligned}
& x^2+y^2-36+k(x-4)=0 \\
& x^2+y^2+k x-4 k-36=0
\end{aligned}
$$
Both circles are intersecting orthogonally, then
$$
\begin{aligned}
-4 k & -36-36=0 \\
-4 k & =72 \\
k & =-18
\end{aligned}
$$
So, equation of required circle
$$
\begin{aligned}
& x^2+y^2-18 x-4(-18)-36=0 \\
& \Rightarrow x^2+y^2-18 x+72-36=0 \\
& \Rightarrow \quad x^2+y^2-18 x+36=0
\end{aligned}
$$
$$
\begin{aligned}
& x^2+y^2=36 \\
& \Rightarrow \quad x^2+y^2-36=0 \\
&
\end{aligned}
$$
and radical axis of two circle is $x-4=0$
So, equation of other circle is
$$
\begin{aligned}
& x^2+y^2-36+k(x-4)=0 \\
& x^2+y^2+k x-4 k-36=0
\end{aligned}
$$
Both circles are intersecting orthogonally, then
$$
\begin{aligned}
-4 k & -36-36=0 \\
-4 k & =72 \\
k & =-18
\end{aligned}
$$
So, equation of required circle
$$
\begin{aligned}
& x^2+y^2-18 x-4(-18)-36=0 \\
& \Rightarrow x^2+y^2-18 x+72-36=0 \\
& \Rightarrow \quad x^2+y^2-18 x+36=0
\end{aligned}
$$
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