# How do you write an equation of a circle with Center: (6, 7) Point on Circle: (13 4)?

Jan 17, 2017

${x}^{2} + {y}^{2} - 12 x - 14 y + 27 = 0$

#### Explanation:

Let say $x \mathmr{and} y$ is another point on the circle.

${\left(x - 6\right)}^{2} + {\left(y - 7\right)}^{2} = {\left(13 - 6\right)}^{2} + {\left(4 - 7\right)}^{2}$
${x}^{2} - 12 x + 36 + {y}^{2} - 14 y + 49 = {\left(7\right)}^{2} + {\left(- 3\right)}^{2}$
${x}^{2} - 12 x + 36 + {y}^{2} - 14 y + 49 = 49 + 9$
${x}^{2} + {y}^{2} - 12 x - 14 y + 36 + 49 - 49 - 9 = 0$
${x}^{2} + {y}^{2} - 12 x - 14 y + 27 = 0$

Jan 17, 2017

${x}^{2} + {y}^{2} - 12 x - 14 y + 27 = 0.$

#### Explanation:

The Centre $C$ of the Circle is $C \left(6 , 7\right)$

The Point $P \left(13 , 4\right)$ is on the Circle. Therefore, the Radius $r$ of the

Circle is distance $C P$. Using, the Distance Formula,

${r}^{2} = C {P}^{2} = {\left(13 - 6\right)}^{2} + {\left(4 - 7\right)}^{2} = 49 + 9 = 58.$

We know that the eqn. of a Circle having Centre at #(h,k) and Radius

$r \text{ } i s : {\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

Therefore, the eqn. of the circle is

${\left(x - 6\right)}^{2} + {\left(y - 7\right)}^{2} = 58 , i . e . ,$

${x}^{2} + {y}^{2} - 12 x - 14 y + 27 = 0$