Find the Circle Through (-10,6) with Center (4,-8) (4,-8) , (-10,6)

$(4, - 8)$ , $(- 10, 6)$

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 $(4, - 8)$ and $(- 10, 6)$ .

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.

$r = \sqrt{{((- 10) - 4)}^{2} + {(6 - (- 8))}^{2}}$

Simplify.

Subtract $4$ from $- 10$ .

$r = \sqrt{{(- 14)}^{2} + {(6 - (- 8))}^{2}}$

Raise $- 14$ to the power of $2$ .

$r = \sqrt{196 + {(6 - (- 8))}^{2}}$

Multiply $- 1$ by $- 8$ .

$r = \sqrt{196 + {(6 + 8)}^{2}}$

Add $6$ and $8$ .

$r = \sqrt{196 + 14^{2}}$

Raise $14$ to the power of $2$ .

$r = \sqrt{196 + 196}$

Add $196$ and $196$ .

$r = \sqrt{392}$

Rewrite $392$ as $14^{2} \cdot 2$ .

Factor $196$ out of $392$ .

$r = \sqrt{196 (2)}$

Rewrite $196$ as $14^{2}$ .

$r = \sqrt{14^{2} \cdot 2}$

Pull terms out from under the radical.

$r = 14 \sqrt{2}$

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, $r = 14 \sqrt{2}$ and the center point is $(4, - 8)$ . The equation for the circle is ${(x - (4))}^{2} + {(y - (- 8))}^{2} = {(14 \sqrt{2})}^{2}$ .

${(x - (4))}^{2} + {(y - (- 8))}^{2} = {(14 \sqrt{2})}^{2}$

Step 3

The circle equation is ${(x - 4)}^{2} + {(y + 8)}^{2} = 392$ .

${(x - 4)}^{2} + {(y + 8)}^{2} = 392$

Step 4

Find the Circle Through (-10,6) with Center (4,-8) (4,-8) , (-10,6)

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