In `DeltaABC`, equation of side BC is x+y-6=0, also the circumcentre and orhtocentre are (3, 1) and (2, 2) respectively, then the equation of the circumcircle of `DeltaABC` is

Options

A. `x^(2)+y^(2)-6x-2y+10=0`
B. `x^(2)+y^(2)-6x-2y=0`
C. `x^(2)+y^(2)-2x-3y-10=0`
D. `x^(2)+y^(2)-2x-3y=0`

Answer

B

Explanation

The circumcircle of `DeltaABC` passes through the reflection of the orthocentre(2, 2) in the line mirror x+y-6=0 i.e. BC. Let `(x_(1), y_(1))` be the reflection of (2, 2) in x+y-6=0. Then,
`(x_(1)-2)/(1)=(y_(1)-2)/(1)=(-2(2+2-6))/(1+1) rArr x_(1)=4, y_(1)=4`
Hence, the equation of the circumcircle, is
`(x-3)^(2)+(y-1)^(2)=(3-4)^(2)+(1-4)^(2)`
or, `x^(2)+y^(2)-6x-2y=0`