The polar of \((1,-2)\) with respect to \(x^2+y^2-10 x-10 y+25=0\) is
The polar of \((1,-2)\) with respect to \(x^2+y^2-10 x-10 y+25=0\) is
\(4 x+7 y+30=0\)
\(4 x+7 y-30=0\)
\(4 x-7 y+30=0\)
\(x+y=0\)
Solution
Given circle is,
\(x^2+y^2-10 x-10 y+25=0\)
Given point ' \(P\) ' is \((1,-2)\)
Equation of polar of point \(P\) with respect to circle is,
\(\begin{aligned}
& x \cdot 1+y(-2)-10\left(\frac{x+1}{2}\right)-10\left(\frac{y-2}{2}\right)+25=0 \\
& \Rightarrow \quad x-2 y-5(x+1)-5(y-2)+25=0 \\
& \Rightarrow \quad x-2 y-5 x-5-5 y+10+25=0 \\
& \Rightarrow \quad-4 x-7 y+30=0 \\
& \Rightarrow \quad 4 x+7 y-30=0 \\
\end{aligned}\)