# Circle has the equation x^2+y^2-6x+10y-15=0, how do you graph the circle using the center (h,k) radius r?

Jul 9, 2018

See explanation...

#### Explanation:

Given:

${x}^{2} + {y}^{2} - 6 x + 10 y - 15 = 0$

Rearrange into standard form for the equation of a circle as follows...

Complete the square for both $x$ and $y$:

$0 = {x}^{2} + {y}^{2} - 6 x + 10 y - 15$

$\textcolor{w h i t e}{0} = {x}^{2} - 6 x + 9 + {y}^{2} + 10 y + 25 - 49$

$\textcolor{w h i t e}{0} = {\left(x - 3\right)}^{2} + {\left(y + 5\right)}^{2} - {7}^{2}$

Adding ${7}^{2}$ to both ends and transposing, this becomes:

${\left(x - 3\right)}^{2} + {\left(y + 5\right)}^{2} = {7}^{2}$

or if you prefer:

${\left(x - 3\right)}^{2} + {\left(y - \left(- 5\right)\right)}^{2} = {7}^{2}$

which is in the form:

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

• the equation of a circle with centre $\left(h , k\right) = \left(3 , - 5\right)$ and radius $r = 7$.

graph{(x^2+y^2-6x+10y-15)((x-3)^2+(y+5)^2-0.08)=0 [-13.67, 17.33, -13, 2.5]}

Jul 9, 2018

The center is $= \left(3 , - 5\right)$ and the radius is $= 7$

#### Explanation:

The standard equation of a circle, center $C = \left(h , k\right)$ and radius $= r$ is

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

Convert the given equation to the standard form by completing the square

${x}^{2} + {y}^{2} - 6 x + 10 y - 15 = 0$

${x}^{2} - 6 x + {y}^{2} + 10 y = 15$

${x}^{2} - 6 x + 9 + {y}^{2} + 10 y + 25 = 15 + 9 + 25$

${\left(x - 3\right)}^{2} + {\left(y + 5\right)}^{2} = 49 = {7}^{2}$

The center is $= \left(3 , - 5\right)$ and the radius is $= 7$

The graph is as follows :

graph{x^2+y^2-6x+10y-15=0 [-14.8, 17.23, -12.3, 3.72]}