I have a circle with equation $(x-a)^2+(y-b)^2=r^2$ and a point $P=(x_1,y_1)$ outside the circle. I am asked to find the equation of the line connecting the tangents from $P$ to the circle (polar line).
I set up a system of equations, but it was too long and very tedious. Any other way? I think the answer is $(x_1-a)(x-a)+(y_1-b)(y-b)=r^2$.
You are being asked to find yhe equation of polar of the circle about thr pole $P(h,k)$. The circle is $f(x,y)=x^2+y^2-2ax-2by+a^2+b^2-r^2=0$ its ploar about $P(h,k)$ will be given by $$hx+ky-a(x+h)-b(y+k)+a^2+b^2-r^2=0. f(h,k) \ne 0 $$
The polar becomes the tangent at $(h,k)$ if $f(h,k)=0$.