Question

How do you write an equation of a circle that passes through the points (3,6), (-1,-2), and (6,5)?

Answer 1

`x^2+y^2+4x-12y-25=0`

Explanation:

`x^2+y^2+2gx+2fy+c=0`

`9+36+6g+12f+c=0`
`6g+12f+c+45=0.....1`
`1+4-2g-4f+c=0`
`-2g-4f+c+5=0.....2`
`36+25+12g+10f+c=0`
`12g+10f+c+61=0....3`

by solving we get g=2, f=-6 c=-25
therefore the equation is `x^2+y^2+4x-12y-25=0`

Answer 2

`x^2+y^2-6*x-2*y-15=0`

Explanation:

This approach requires solving a system of three simultaneous first-degree equations.

Let the equation of the circle in a `x,y` plane be

`x^2+y^2+a*x+b*y+c=0`

where `a`, `b`, and `c` are unknowns.
Construct three equations about `a`, `b`, and `c`, one for each point given:
`3^2+6^2+3*a+6*b+c=0`,
`(1)^2+(-2)^2+(-1)*a+(-2)*b+c=0`, and
`6^2+5^2+6*a+5*b+c=0`

Solving for the system shall give
`a=-6`, `b=-2`, and `c=-15`

Thus the Equation of the circle:
`x^2+y^2-6*x-2*y-15=0`

Reference:
"[The] Equation of [a] circle passing through 3 given points", Maths Department, Queen's College, http://www.qc.edu.hk/math/Advanced%20Level/circle%20given%203%20points.htm