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`

Options

A. `5x^(2)+5y^(2)+8x -14y-16=0`
B. `5x^(2)+5y^(2)-8x-14y-32=0`
C. `5x^(2)+5y^(2)-8x+14y-4=0`
D. `5x^(2)+5y6(2)+8x+14y+12=0`

Answer

B

Explanation

The given circle `x^(2)+y^(2)-4x-6y-12=0` has its centre at (2, k) and radius equal to 5.
Let (h, k) be the centre of the required circle.
Then, the point (h, k) divides the line joining (-1, -1) to (2, 3) in the ratio 3:2, where 3 is the radius of the required circle. Thus,
we have
`h=(3xx2+2xx(-1))/(3+2)=(4)/(5) and k=(3xx3+2xx(-1))/(3+2)=(7)/(5)`
Hence, the equation of the reuired circle is
`(x-(4)/(5))^(2)+(y-(7)/(5))^(2)=3^(2)`
`rArr 5x^(2)+5y^(2)-8x-14y-32=0`