The locus of centre of circles which cuts orthogonally the circle x^2+y^2-4 x+8=0 and touches x+1=0 , is

Step-by-step solution

Let equation of circle be x^2+y^2+2 g x+2 f y+c=0 Given equation of circle is x^2+y^2-4 x+8=0 The centres of above circles are (-g,-f) and (2,0) . Condition of orthogonality is 2 (g₁ g₂+f₁ f₂ )=c₁+c₂ 2(g (-2)+(f) 0)=c+8 -4 g=c+8 ....(i) Also, the assume circle touch the line x+1=0 . The perpendicular drawn from centre to the line is equal to radius. -g+1 1^2 = g^2+f^2-c -g+1= g^2+f^2-c On squaring both sides, we get g^2+1-2 g=g^2+f^2-c c=f^2+2 g-1 Putting the value of c in Eq. (i), we get -4 g=f^2+2 g-1+8 f^2+2 g+4