If one of the diameters of the circle, given by the equation, `x^2+y^2-4x+6y-12=0` , is a chord of a circle S, whose centre is at `(-3,""2)` , then the radius of S is : (1) `5sqrt(2)` (2) `5sqrt(3)` (3) 5 (4) 10

Options

A. `5sqrt(3)`
B. 5
C. 10
D. `5sqrt(2)`

Answer

A

Explanation

Let PQ be a diameter of the circle `x^(2)+y^(2)-4x+6y-12=0`. Then,
`PQ=2sqrt(2^(2)+(-3)^(2)+12)=10`
`:. PR = RQ=5`
In right triangle PRC, we obtain
`CP^(2)=CR^(2)+PR^(2)`
`rArr CP^(2)=25+(2+3)^(2)+(-3-2)^(2)`
`rArr CP^(2)=25+50=75`
`rArr CP=5sqrt(3)`