Find the Circle Through (-2,10) with Center (-2,-3) (-2,-3) , (-2,10)

$(- 2, - 3)$ , $(- 2, 10)$

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 $(- 2, - 3)$ and $(- 2, 10)$ .

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.

$r = \sqrt{{((- 2) - (- 2))}^{2} + {(10 - (- 3))}^{2}}$

Simplify.

Multiply $- 1$ by $- 2$ .

$r = \sqrt{{(- 2 + 2)}^{2} + {(10 - (- 3))}^{2}}$

Add $- 2$ and $2$ .

$r = \sqrt{0^{2} + {(10 - (- 3))}^{2}}$

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

$r = \sqrt{0 + {(10 - (- 3))}^{2}}$

Multiply $- 1$ by $- 3$ .

$r = \sqrt{0 + {(10 + 3)}^{2}}$

Add $10$ and $3$ .

$r = \sqrt{0 + 13^{2}}$

Raise $13$ to the power of $2$ .

$r = \sqrt{0 + 169}$

Add $0$ and $169$ .

$r = \sqrt{169}$

Rewrite $169$ as $13^{2}$ .

$r = \sqrt{13^{2}}$

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

$r = 13$

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, $r = 13$ and the center point is $(- 2, - 3)$ . The equation for the circle is ${(x - (- 2))}^{2} + {(y - (- 3))}^{2} = {(13)}^{2}$ .

${(x - (- 2))}^{2} + {(y - (- 3))}^{2} = {(13)}^{2}$

Step 3

The circle equation is ${(x + 2)}^{2} + {(y + 3)}^{2} = 169$ .

${(x + 2)}^{2} + {(y + 3)}^{2} = 169$

Step 4

Find the Circle Through (-2,10) with Center (-2,-3) (-2,-3) , (-2,10)

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