Find the equation of circle of a circle which passes through the points `(0,-6), (1,1) and (7,-7)`.

Explanation

Let the equation of circle be
`x^(2)+y^(2)+2gx+2fy+c=0`....(1)
It passes through the point `(0, -6)`
`0+36+0-23f+c=0`
`rArr-12f+c=-36`....(2)
It passes through the point (1, 1)
`1+1+2g_+2f+c=0`
`rArr2g+2f+c=-2`...(3)
It pass through the point (7, -7)
`49+49 +14g -14f+c=0`
`rArr 14g -14f +x = -98` ...(4)
Subtract eq. (3) from eq. (2) and eq. (4) frm eq. (3)
`-2g - 14f = 34` ....(5)
and `-12g +16f = 96` ....(6)
Multiply eq. (5) by 6 and subtract eq. (6) from it
`- 12g -84f = - 204`
On subtracting `(-12g+16f=96)/(-100f=-300)`
`rArrf=3`
From eq. (2)
`-36 +c = -36`
`rArr c=0`
and from eq. (5)
`-2g - 42 = -34`
`rArr g=-4`
`:.` From eq. (1), the equation of circle
`x^(2)+y^(2) -8x +6y =0`.