Question

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
A. `5sqrt2`
B. `5sqrt3`
C. 5
D. 10

Answer 1

Correct Answer - B
Given equation of a circle is `x^(2)+y^(2)-4x + 6y -12 =0`, whose centre is (2, - 3) and radius
`=sqrt(2^(2)+(-3)^(2)+12)=sqrt(4+9+12)=5`
Now, according to given information, we have the following figure.
`x^(2)+y^(2)-4x + 6y-12=0`
Clearly, `AO bot BC, as O` is mid-point of the chord
Now, in `DeltaAOB`, we have
`OA =sqrt((-3-2)^(2)+(2+3)^(2))`
`=sqrt(25+25)=sqrt50=5sqrt2`
`and OB = 5`
`AB=sqrt(OA^(2)+OB^(2))=sqrt(50+25)=sqrt75=5sqrt3`