In triangle `A B C` , the equation of side `B C` is `x-y=0.` The circumcenter and orthocentre of triangle are (2, 3) and (5, 8), respectively. The equation of the circumcirle of the triangle is a) `x^2+y^2-4x+6y-27=0` b) `x^2+y^2-4x-6y-27=0` c) `x^2+y^2+4x-6y-27=0` d) `x^2+y^2+4x+6y-27=0`

Options

A. `x^(2)+y^(2)-4x-6y-27=0`
B. `x^(2)+y^(2)-4x-6y-36=0`
C. `x^(2)+y^(2)-4x-6y-24=0`
D. x^(2)+y^(2)-4x-6y-15=0`

Answer

2

Explanation

Let the orthocenter be `H(5,8)`
Now, `/_ HBM =(pi)/(2)-C`
Also, `/_ DBC=/_ DAC =(pi)/(2)-C`
Hence, `Delta CMH` and `Delta BMD` are congruent. Therefore,
`HM=MD`
D is the image of H in the line `x-y=0`, which is `D(8,5)`.
Thus, the equation of the circumcircle is
`(x-2)^(2)+(y-3)^(2)=(8-2)^(2)+(5-3)^(2)`
i.e., `x^(2)+y^(2)-4x-6y-6y-27=0`