Question

Find the equation of the circle whose radius 3 and which touches internally the circle $x^2+y^2-4x-6y-12=0$ at the point (-1, -1)

Answer 1

$x^2+y^2$ 4x 6y 12 = 0
Centre A is (2, 3) and radius 5 = PA, B (h, k) is the centre of the required circle of radius BP = 3 which touches the given circle internally at P (- 1, - 1)
$\therefore$ BA = PA PB = 5 3 = -2
Thus B divides PA in the ratio 3 : 2.
$\therefore h=\frac{3.2+2(-1)}{3+2}=\frac{4}{5},k=\frac{3.2+2(-1)}{3+2}=\frac{7}{5}$
Hence the required circle is
(x-4/5)${}^2$ + (y - 7/5)${}^2$ =3${}^2$
Or $5x^2+5y^2$ - 8x - 14y 32 = 0.