If the circle S =0 cuts the circles x^2+y^2-2 x+6 y=0 . x^2+y^2-4 x-2 y+6=0 and x^2+y^2-12 x+2 y+3=0 orthogonally, then equation of the tangent at (0,3) on S= 0 is

Step-by-step solution

aligned Let & S₁ x^2+y^2-2 x+6 y=0 & S₂ x^2+y^2-4 x-2 y+6=0 aligned and S₃ x^2+y^2-12 x+2 y+3=0 Also let S x^2+y^2+2 g x+2 f y+c=0 Since, S₁ and S cut orthogonally So, 2 g₁ g₂+2 f₁ f₂=c₁+c₂ 2 g (-1)+2 f (3)=0+c -2 g+6 f=c...(i) Similarly S₂ and S 2 g(-2)+2 f(-1)=6+c -4 g-2 f=6+c..(ii) and S₃ and S 2 g(-6)+2 f(1)=3+c -12 g+2 f=3+c...(iii) After solving (i), (ii) and (iii), we get g=0, f= -3 4 and c= -9 2 Since, equation of tangent of S 0 is aligned & x x₁+y y₁+g (x+x₁ )+f (y₁+y )+c=0 & x 0+3 y+0- 3 4 (3+y)- 9 2 =0 &