Find the equation of the circle if the chord of the circle joining (1, 2) and `(-3,1)` subtents `90^0` at the center of the circle.

Explanation

The chord joining points A(1,2) and B(-3,1) subtends `90^(@)` at the center of the circle. Then AB subtend an angle `45^(@)` at point C on the circumference. Therefore, the equation of the circle is
`(x-1)(x+3)+(y-2)(y-1)`
`= +-cot 45^(@)[(y-2)(x+3)-(x-1)(y-1)]`
or `x^(2)+y^(2)+2x-3y-1= +-[4y-x-7]`
or `x^(2)+y^(2)+3x-7y+6=0` and `x^(2)+y^(2)+x+y-8=0`