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The locus of centre of a circle which passes through the origin and cuts off a length of 4 unit from the line $x=3$ is
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Verified Answer
The correct answer is:
$y^2+6 x=13$
Let centre of circle be $C(-g,-f)$, then equation of circle passing through origin be
$x^2+y^2+2 g x+2 f y=0$
$\begin{array}{lc}
\therefore \text { Distance, } d=|-g-3|=g+3 \\
\text { In } \triangle A B C, & (B C)^2=A C^2+B A^2 \\
\Rightarrow & g^2+f^2=(g+3)^2+2^2 \\
\Rightarrow & g^2+f^2=g^2+6 g+9+4 \\
\Rightarrow & f^2=6 g+13
\end{array}$
Hence, required locus is $y^2+6 x=13$
$x^2+y^2+2 g x+2 f y=0$
$\begin{array}{lc}
\therefore \text { Distance, } d=|-g-3|=g+3 \\
\text { In } \triangle A B C, & (B C)^2=A C^2+B A^2 \\
\Rightarrow & g^2+f^2=(g+3)^2+2^2 \\
\Rightarrow & g^2+f^2=g^2+6 g+9+4 \\
\Rightarrow & f^2=6 g+13
\end{array}$
Hence, required locus is $y^2+6 x=13$
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