# How do you write an equation of a circle with Center (5, -6) Point on Circle: (18 -6)?

Nov 23, 2016

The equation of the circle can be written:

${\left(x - 5\right)}^{2} + {\left(y + 6\right)}^{2} = 169$

or

${x}^{2} + {y}^{2} - 10 x + 12 y - 147 = 0$

#### Explanation:

The distance between $\left(5 , - 6\right)$ and $\left(18 , - 6\right)$ is $18 - 5 = 13$, since both lie on the horizontal line $y = - 6$.

If the points did not lie on the same horizontal or vertical line then you would use the Pythagorean distance formula:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

but it would give the same answer:

$d = \sqrt{{\left(18 - 5\right)}^{2} + {\left(\left(- 6\right) - \left(- 6\right)\right)}^{2}} = \sqrt{{13}^{2} + {0}^{2}} = 13$

Thus the radius of the circle is $13$.

The equation of a circle with centre $\left(h , k\right)$ and radius $r$ can be written:

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

So in the given example, we can write:

${\left(x - 5\right)}^{2} + {\left(y - \left(- 6\right)\right)}^{2} = 13$

which simplifies to:

${\left(x - 5\right)}^{2} + {\left(y + 6\right)}^{2} = 169$

If you prefer, this can be expanded to:

${x}^{2} - 10 x + 25 + {y}^{2} + 12 y + 36 = 169$

then simplified to:

${x}^{2} + {y}^{2} - 10 x + 12 y - 147 = 0$