Find the Circle Through (-3,-2) with Center (0,-2) (-3,-2) , (0,-2)

$(- 3, - 2)$ , $(0, - 2)$

Step 1

Find the radius $r$ for the circle. The radius is any line segment from the center of the circle to any point on its circumference. In this case, $r$ is the distance between $(0, - 2)$ and $(- 3, - 2)$ .

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Substitute the actual values of the points into the distance formula.

$r = \sqrt{{((- 3) - 0)}^{2} + {((- 2) - (- 2))}^{2}}$

Simplify.

Subtract $0$ from $- 3$ .

$r = \sqrt{{(- 3)}^{2} + {((- 2) - (- 2))}^{2}}$

Raise $- 3$ to the power of $2$ .

$r = \sqrt{9 + {((- 2) - (- 2))}^{2}}$

Multiply $- 1$ by $- 2$ .

$r = \sqrt{9 + {(- 2 + 2)}^{2}}$

Add $- 2$ and $2$ .

$r = \sqrt{9 + 0^{2}}$

Raising $0$ to any positive power yields $0$ .

$r = \sqrt{9 + 0}$

Add $9$ and $0$ .

$r = \sqrt{9}$

Rewrite $9$ as $3^{2}$ .

$r = \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$r = 3$

Step 2

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the equation form for a circle with $r$ radius and $(h, k)$ as the center point. In this case, $r = 3$ and the center point is $(0, - 2)$ . The equation for the circle is ${(x - (0))}^{2} + {(y - (- 2))}^{2} = {(3)}^{2}$ .

${(x - (0))}^{2} + {(y - (- 2))}^{2} = {(3)}^{2}$

Step 3

The circle equation is ${(x - 0)}^{2} + {(y + 2)}^{2} = 9$ .

${(x - 0)}^{2} + {(y + 2)}^{2} = 9$

Step 4

Simplify the circle equation.

$x^{2} + {(y + 2)}^{2} = 9$

Step 5

Find the Circle Through (-3,-2) with Center (0,-2) (-3,-2) , (0,-2)

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