Question

If the length of tangent from (f,g) to the circle x² +y² =6 be twice the length of the tangent from (f,g) to circle x² +y² +3x+3y = 0 then prove that f² +g² +4f+4g+2 = 0

Answer 1

According to the question

\(\sqrt{f^2+g^2+-6}=2 \sqrt{f^2+g^2+3f+3g}\)

Squaring both side

f² +g² –6 = 4 (f² +g² +3f+3g)

3f² +3g² +12f+12g+6 = 0

Divide by 3 we get f² +g² +4f+4g+2 = 0