The diameters of a circle are along `2x + y -7=0 and x + 3y -11=0`.Then the equation of this circle, whichalso passes through (5, 7) is

Options

A. `x^(2)+y^(2)-4x-6y-16=0`
B. `x^(2)+y^(2)-4x-6y-20=0`
C. `x^(2)+y^(2)-4x-6y-12=0`
D. `x^(2)+y^(2)+4x+6y-12=0`

Answer

C

Explanation

The intersection point of diameter lines is (2,3) which is the centre of circlee.
Now, radius`=sqrt((5-2)^(2)+(7-3)^(2))`
`=sqrt(9+16)=5`
Hence, requried equation of circle is
`(x-2)^(2)+(y-3)^(2)=5^(2)`
`impliesx^(2)+y^(2)-4x-6y-12=0`