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

Answer

`5x^(2)+5y^(2)-8x-14y-32=0`

Explanation

The given circle is
`x^(2)+y^(2)-4x-6y-12=0`
whose center is `C_(1)(2,3)` and radius `r_(1)` is `C_(1)A=5`.
If `C_(2)(h,k)` is the centre of the circle of radius 3 which touches the circle (1) internally at `A(-1,-1)`, then `C_(2)A=3` and `C_(1)C_(2)=C_(1)A-C_(2)A=5-3=2`.
Thus, `C_(2)(h,k)` divides `C_(1)` A in the ratio `2:3` internally. Therefore,
`h=(2(-1)+3xx2)/(2+3)=(4)/(5)`
and`k=(2(-1)+3xx3)/(2+3)=(7)/(5)`
Hence, the equaiton of the required circle is `(x-4//5)^(2)+(y-7//5)^(2)=3^(2), i.e., 5x^(2)+5y^(2)-8x-14y-32=0`.