Consider the equation, x2+y2−4x−6y−12=0 The center of above is C1(2,3) and radius is, r1‌‌=√4+9+12 ‌‌=5 Let the center of required circle be C2(h,k) and radius is given as 3 touches given circle as A(−1,−1) The point A(−1,−1) divides the line joining the centers C1(2,3) and C2(h,k) externally in 5: 3 so, (−1,−1)‌‌=(‌
5h−3(2)
5−3
,‌
5k−3(3)
5−3
) ‌‌=(‌
5h−6
2
,‌
5k−9
2
) From above, 5h−6=−2‌ and ‌5k−9=−2 This implies, h=‌