Let centre of circle C is C1 Circle C:x2+y2+4x−6y−12=0 With centre C2:(−2,3),r=5 Tangent at M(1,−1) to the circle C is xx1+yy1+4(
x+x1
2
)−6(
y+y1
2
)−12=0 x−y+2(x+1)−3(y−1)−12=0 ⇒3x−4y−7=0 So, equation of circle C touching the line 3x−4y−7=0 at (1,−1) (x−1)2+(y+1)2+λ(3x−4y−7)=0 It passes through (4,0) (4−1)2+(0+1)2+λ(12−0−7)=0 ⇒9+1+5λ=0 ⇒λ=−2 So equation of C:(x−1)2+(y+1)2−2(3x−4y−7)=0 x2+y2−8x+10y+16=0 Centre C1=(4,−5) Radius of C=√16+25−16=5