Find the co - ordinates of the circles $x^{2} + y^{2} - 4 x - 2 y = 4$ and $x^{2} + y^{2} - 12 x - 8 y = 12$ touch each other. Also find equations of common tangents touching the circles in distinct points.

Question

Q. The coordinates of the point at which the circles

x^{2} + y^{2} - 4 x - 2 y - 4 = 0

and

x^{2} + y^{2} - 12 x - 8 y - 36 = 0

touch each other, are

Answer 1

C_{1} (2, 1), r_{1} = 3, C_{2} (6, 4), r = 8, C_{1} C_{2} = 5 = r_{2} - r_{1}

Answer 2

Hence the circles touch internally. The common tangent being

Answer 3

S_{1} - S_{2}

= 0 or 4x + 3y + 4 =0.

Answer 4

and common normal

C_{1} C_{2}

is 3x - 4y - 2 = 0

Answer 5

Solving these, the point of contact is

(- \frac{2}{5}, - \frac{4}{5})

Answer 6

The point of contact can also be found by ratio formula. Since the circle touch internally there will be no direct common tangents.

Find the co - ordinates of the circles x2+y2−4x−2y=4 and x2+y2−12x−8y=12 touch each other. Also find equations of common tangents touching the circles in distinct points.