Tangents drawn from the point `P(1,8)` to the circle `x^(2)+y^(2)-6x-4y-11=0` touch the circle at points A and B. The equation of the cricumcircle of triangle PAB is

Options

A. `x^(2)+y^(2)+4x-6y+19=0`
B. `x^(2)+y^(2)-4x-10y+19=0`
C. `x^(2)+y^(2)-2x+6y-29=0`
D. `x^(2)+y^(2)-6x-4y+19=0`

Answer

B

Explanation

Clearly, `anglePAC=anglePBC= 90^(@)`. So, circle passing through P, A and B also passes through C and PC is its diameter. So, its equation is

`(x-1) (x-3) + (y-8) (y-2)=0`
or, `x^(2)+y^(2)-4x-10y+19=0`
`ul(ALITER)` The equation of the common chord AB is
`x+8y-3(x+1)-2(y+8)-11=0 or, x-3y+15=0`
Clearly, the circumcircle of `DeltaPAB` passes through the intersection of circle `x^(2)+y^(2)-6x-4y-11=0` and the chord of contact AB whose equation is x-3y+15=0. So, the equation of the circumcircle is
`x^(2)+y^(2)-6x-4y-11+lambda(x-3y+15)=0 ...(i)`
This passes through P (1, 8).
`:. 16-8lambda=0 rArr lambda = 2`
Putting `lambda = 2 ` in (i), we obtain `x^(2)+y^(2)-4x=10y+19=0`
as the equation of the required circle.