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2023-02-22

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?

Helen Merritt

Beginner2023-02-23Added 3 answers

A two-dimensional number line, such as two perpendicular number lines oraxes, can be thought of as a coordinate system. This coordinate system is typical: The x-axis and y-axis are the names of the horizontal and vertical axes, respectively. The origin is the point at which all lines in a coordinate system intersect.

Riley Cisneros

Beginner2023-02-24Added 7 answers

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

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

Complete the square to change the given equation to standard form.

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

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

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

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

The center is $=(3,-5)$ 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]}

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

Complete the square to change the given equation to standard form.

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

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

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

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

The center is $=(3,-5)$ 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]}