Question

Find the equation of tangent to the circle x^2+y^2-4x+4y-8=0 at (-2,-2) and the normal equation

Answer 1

Given equation of circle x

2

+y

2

+4x−4y+4=0 ....(1)

Center is at (−2,2)

Radius r=

4+4−4

=2

Now, given tangent to eqn (1) makes equal intercepts on the coordinate axes,

So, let the equation of tangent be x±y=a

Now since, length of tangent from the center =Radius

⇒

2

∣−2±2−a∣

=2

Taking (+) sign , we get

⇒a=±2

2

Hence equation of the required tangent is x+y=±2

2

Hope it will help you