Find the Circle Through (-6,6) with Center (0,0) (0,0) , (-6,6)

$(0, 0)$ , $(- 6, 6)$

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 $(0, 0)$ and $(- 6, 6)$ .

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.

$r = \sqrt{{((- 6) - 0)}^{2} + {(6 - 0)}^{2}}$

Simplify.

Subtract $0$ from $- 6$ .

$r = \sqrt{{(- 6)}^{2} + {(6 - 0)}^{2}}$

Raise $- 6$ to the power of $2$ .

$r = \sqrt{36 + {(6 - 0)}^{2}}$

Subtract $0$ from $6$ .

$r = \sqrt{36 + 6^{2}}$

Raise $6$ to the power of $2$ .

$r = \sqrt{36 + 36}$

Add $36$ and $36$ .

$r = \sqrt{72}$

Rewrite $72$ as $6^{2} \cdot 2$ .

Factor $36$ out of $72$ .

$r = \sqrt{36 (2)}$

Rewrite $36$ as $6^{2}$ .

$r = \sqrt{6^{2} \cdot 2}$

Pull terms out from under the radical.

$r = 6 \sqrt{2}$

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the equation form for a circle with $r$ radius and $(h, k)$ as the center point. In this case, $r = 6 \sqrt{2}$ and the center point is $(0, 0)$ . The equation for the circle is ${(x - (0))}^{2} + {(y - (0))}^{2} = {(6 \sqrt{2})}^{2}$ .

${(x - (0))}^{2} + {(y - (0))}^{2} = {(6 \sqrt{2})}^{2}$

The circle equation is ${(x - 0)}^{2} + {(y - 0)}^{2} = 72$ .

${(x - 0)}^{2} + {(y - 0)}^{2} = 72$

Simplify the circle equation.

$x^{2} + y^{2} = 72$

Find the Circle Through (-6,6) with Center (0,0) (0,0) , (-6,6)

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