Number of circles intersecting x^2+y^2=4 , x^2+y^2-2 x-3=0 and x^2+y^2-2 y-3=0 orthogonally is

Step-by-step solution

S₁ x^2+y^2=4 ...(i) S₂ x^2+y^2-2 x-3=0 ...(ii) S₃ x^2+y^2-2 y-3=0 ...(iii) Let S₄ x^2+y^2+2 g x+2 f y+c be the circle which is orthogonal with the circles S₁, S₂ and S₃ . Condition of orthogonal, 2 (g g^ +f f^ )=c+c^ Now, for S₁ and S₄ ; g 0+f 0=c-4 c=4 For S₂ and S₄2[g (-1)+f (0)]=4-3 g=- 1 2 For S₃ and S₄ , 2[g (0)+f (-1)]=4-3 f=- 1 2 Now, radius of S₄ , r^2=g^2+f^2-c= ( 1 4 + 1 4 -4 )=- 7 2 < 0 Thus, no such S₄ circle is possible which is orthogonal to S₁, S₂, S₃ .