Question

The center of the circle ${\left(x+3\right)}^2+{\left(y-2\right)}^2=13$is$\left(3,-2\right)$

Short answer

False

Step-by-step solution

Step 1. Given information

An equation of a circle is$(x+3{)}^2+(y-2{)}^2=13$.

Step 2. State whether the statement is true or false.

The standard form of an equation of a circle $(x-h{)}^2+(y-k{)}^2=r^2$

has a radius $r$and center $\left(h,k\right)$.

Thus, in the equation$(x+3{)}^2+(y-2{)}^2=13$

has the radius $\sqrt{13}$and center $(-3,2)$.

Therefore, the statement is false.

The center is not $(3,-2)$but $(-3,2)$.