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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 circles S, whose centre is at `(-3,2)` , then the radius of S is

A

`5sqrt(3)`

B

5

C

10

D

`5sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
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)`
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