If the line `2x-y+1=0` touches the circle at the point `(2,5)` and the centre of the circle lies in the line `x+y-9=0.` Find the equation of the circle.

Options

A. `x^(2)+y^(2)-12x+6y+5=0`
B. `x^(2)+y^(2)-12x-6y-5=0`
C. `x^(2)+y^(2)+12x+6y+15=0`
D. `x^(2)+y^(2)-12x-6y+25=0`

Answer

D

Explanation

Clearly, the centre of the circle is the point of intersection of the line x+y=9 and the normal at P i.e. a line passing through P and perpendicular to the tangent at P.
The equation of CP is
`y-5=-(1)/(2)(x-2)rArrx+2y=12 " " ...(i)` Solving this equation with x+y=9, we obtain that the coordinates of C are (6, 3).

Now,
Radius of the required circle
`=CP=sqrt((6-2)^(2)+(5-3)^(2))s=sqrt(20)` .
Hence, the equation of the required circle is
`(x-6)^(2)+(y-3)^(2)=20 " or ", x^(2)+y^(2)-12x-6y+25=0`