# 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?

##### 2 Answers

See explanation...

#### Explanation:

Given:

#x^2+y^2-6x+10y-15=0#

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

Complete the square for both

#0 = x^2+y^2-6x+10y-15#

#color(white)(0) = x^2-6x+9+y^2+10y+25-49#

#color(white)(0) = (x-3)^2+(y+5)^2-7^2#

Adding

#(x-3)^2+(y+5)^2=7^2#

or if you prefer:

#(x-3)^2+(y-(-5))^2=7^2#

which is in the form:

#(x-h)^2+(y-k)^2=r^2#

- the equation of a circle with centre
#(h, k) = (3, -5)# 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]}

The center is

#### Explanation:

The standard equation of a circle, center

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

The center is

The graph is as follows :

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