If the circles x2+y2−4x−6y−12=0 and 5(x2+y2)−8x−14y−32=0 – InterviewSolution

1.	If the circles x2+y2−4x−6y−12=0 and 5(x2+y2)−8x−14y−32=0 touch each other then their point of contact is
Answer» If the circles $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ and $5 (x^{2} + y^{2}) - 8 x - 14 y - 32 = 0$ touch each other then their point of contact is

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