Find the Circle Through (-6,-1) with Center (2,7) (2,7) , (-6,-1)

$(2, 7)$ , $(- 6, - 1)$

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 $(2, 7)$ and $(- 6, - 1)$ .

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.

$r = \sqrt{{((- 6) - 2)}^{2} + {((- 1) - 7)}^{2}}$

Simplify.

Subtract $2$ from $- 6$ .

$r = \sqrt{{(- 8)}^{2} + {((- 1) - 7)}^{2}}$

Raise $- 8$ to the power of $2$ .

$r = \sqrt{64 + {((- 1) - 7)}^{2}}$

Subtract $7$ from $- 1$ .

$r = \sqrt{64 + {(- 8)}^{2}}$

Raise $- 8$ to the power of $2$ .

$r = \sqrt{64 + 64}$

Add $64$ and $64$ .

$r = \sqrt{128}$

Rewrite $128$ as $8^{2} \cdot 2$ .

Factor $64$ out of $128$ .

$r = \sqrt{64 (2)}$

Rewrite $64$ as $8^{2}$ .

$r = \sqrt{8^{2} \cdot 2}$

Pull terms out from under the radical.

$r = 8 \sqrt{2}$

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, $r = 8 \sqrt{2}$ and the center point is $(2, 7)$ . The equation for the circle is ${(x - (2))}^{2} + {(y - (7))}^{2} = {(8 \sqrt{2})}^{2}$ .

${(x - (2))}^{2} + {(y - (7))}^{2} = {(8 \sqrt{2})}^{2}$

Step 3

The circle equation is ${(x - 2)}^{2} + {(y - 7)}^{2} = 128$ .

${(x - 2)}^{2} + {(y - 7)}^{2} = 128$

Step 4

Find the Circle Through (-6,-1) with Center (2,7) (2,7) , (-6,-1)

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