The equation obtained by transforming $x^2+y^2-6 x+10 y-2=0$ to the parallel axis through $(3,-5)$ is
The equation obtained by transforming $x^2+y^2-6 x+10 y-2=0$ to the parallel axis through $(3,-5)$ is
$x^2+y^2=16$
$x^2+y^2=9$
$x^2+y^2=25$
$x^2+y^2=36$
Solution
Given equation is
$
\begin{aligned}
x^2+y^2-6 x+10 y-2 & =0 \\
(x-3)^2+(y+5)^2 & =36
\end{aligned}
$
The equation obtained by transforming
$(x-3)^2+(y+5)^2=36$ to parallel axis through $(3,-5)$ is
Put $x-3=h$ and $y+5=k$
$
\therefore h^2+k^2=36
$
i.e. $x^2+y^2=36$