Find the equation of a circle which cuts the circle x^2+y^2-6 x+4 y-3=0 orthogonally, while passing through (3,0) and touching the Y -axis.

Step-by-step solution

When two circles intersects each other orthogonally, then 2 (g₁ g₂+f₁ f₂ )=c₁ c₂ , where two circles are aligned & x^2+y^2+2 g₁ x+2 f₁ y+c₁=0 and & x^2+y^2+2 g₂ x+2 f₂ y+c₂=0 aligned Let c(h, k) be the centre of required circle which passes through (3,0) and also touches Y -axis? Radius = (h-3)^2+(k-0)^2 =|h| (h-3)^2+k^2=h^2 k^2-6 h+9=0 ...(i) Required circle (x-h)^2+(y-k)^2=h^2 x^2+y^2-2 h x-2 k y+k^2=0 ...(ii) g₁=-h₁, f₁=-k, c₁=k^2 Circle (ii) is intersected orthogonally by x^2+y^2-6 x+4 y-3=0 g₂=-3, f₂=2, c₂=-3