Question

For the following exercises, write the standard form of the equation of the circle with the given center with point on the circle.

$Center(3,-2)withpoint(3,6)$

Short answer

The standard form of the equation of the circle is${(x-3)}^2+{(y+2)}^2=64.$

Step-by-step solution

Step 1. Given Information

The given center is $(3,-2)$and point on the circle is$(3,6).$

We have to write the standard form of the equation of the circle with the given center point on the circle.

Step 2. Use the Distance Formula to find the radius

The radius is the distance from the center to any point on the circle so we can use the distance formula to calculate it.

The distance formula is $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}.$

Now, substitute the values as

$$\begin{gathered} (x_1,y_1)=(3,-2) \\ (x_2,y_2)=(3,6) \end{gathered}$$

So,

$$\begin{gathered} r=\sqrt{{(3-3)}^2+{(6+2)}^2} \\ r=\sqrt{0+64} \\ r=8 \end{gathered}$$

Step 3. Use the standard form of the equation of a circle

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

Now, substitute the values $r=8$and $h=3,k=-2$

${(x-3)}^2+{(y+2)}^2=64$