Chapter T, Problem 3BDT

To determine

To find: The center and radius of the circle with equation $x^2+y^2-6x+10y+9=0$.

Answer to Problem 3BDT

The center of the circle is at the point $\left(3,-5\right)$ and the radius of the circle is 5.

Explanation of Solution

Formula used:

The equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where the center of the circle is at the point $\left(h,k\right)$, r is the radius of the circle.

Calculation:

It is given that the equation of the circle is $x^2+y^2-6x+10y+9=0$.

Simplify the above equation as follows,

$\begin{array}{l}{x}^2+y^2-6x+10y+9=0\\ x^2-6x+9+y^2+10y+25-25=0\\ {\left(x-3\right)}^2+{\left(y+5\right)}^2=25\\ {\left(x-3\right)}^2+{\left(y+5\right)}^2=5^2\end{array}$

Compare the equations ${\left(x-3\right)}^2+{\left(y+5\right)}^2=5^2$ and ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$.

It is clear that $h=3$, $k=-5$, and $r=5$.

The center of the circle is at the point $\left(h,k\right)$, that is $\left(3,-5\right)$.

The radius of the circle is $r=5$.

Thus, the center of the circle is at the point $\left(3,-5\right)$ and the radius of the circle is 5.