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

Apr 5, 2018

${x}^{2} + {y}^{2} + 4 x - 12 y - 25 = 0$

#### Explanation:

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

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

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

Apr 5, 2018

${x}^{2} + {y}^{2} - 6 \cdot x - 2 \cdot y - 15 = 0$

#### 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

${x}^{2} + {y}^{2} + a \cdot x + b \cdot y + c = 0$

where $a$, $b$, and $c$ are unknowns.
Construct three equations about $a$, $b$, and $c$, one for each point given:
${3}^{2} + {6}^{2} + 3 \cdot a + 6 \cdot b + c = 0$,
${\left(1\right)}^{2} + {\left(- 2\right)}^{2} + \left(- 1\right) \cdot a + \left(- 2\right) \cdot b + c = 0$, and
${6}^{2} + {5}^{2} + 6 \cdot a + 5 \cdot b + c = 0$

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

Thus the Equation of the circle:
${x}^{2} + {y}^{2} - 6 \cdot x - 2 \cdot y - 15 = 0$

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