Question

The graph of the equation${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center$\left(\text{\_\_\_\_,\_\_\_}\right)$and radius .

Short answer

The graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$ is a circle with center$\left(1,2\right)$ and radius3.

Step-by-step solution

Step 1. Apply the concept of circles.

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

Step 2. Interpret the value of center. 

Consider the provided equation.

${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$

Rewrite the equation as provided below.

${\left(x-1\right)}^2+{\left(y-2\right)}^2=3^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, $\left(h,k\right)$and $\left(1,2\right)$.

Here, $h=1\text{and}k=2$.

Therefore, center of the circle is 2.

Step 3. Interpret the value of radius. 

Consider the provided equation.

${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$

Rewrite the equation as provided below.

${\left(x-1\right)}^2+{\left(y-2\right)}^2=3^2$

Recall that the standard form of the equation of the circle is ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$, where $\left(h,k\right)$ denote the center of the circle and r denote the radius.

Compare, $r^2$and $3^2$.

Here, $r=3$.

Therefore, the radius of the circle is 3.

Thus, the graph of the equation ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$is a circle with a center$\left(1,2\right)$ and radius of 3.