Q 6. - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

False

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 $(-3,2)$.