Find the Circle Through (7,4) with Center (-3,6) (7,4) , (-3,6)

$(7, 4)$ , $(- 3, 6)$

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 $(- 3, 6)$ and $(7, 4)$ .

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.

$r = \sqrt{{(7 - (- 3))}^{2} + {(4 - 6)}^{2}}$

Simplify.

Multiply $- 1$ by $- 3$ .

$r = \sqrt{{(7 + 3)}^{2} + {(4 - 6)}^{2}}$

Add $7$ and $3$ .

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

Raise $10$ to the power of $2$ .

$r = \sqrt{100 + {(4 - 6)}^{2}}$

Subtract $6$ from $4$ .

$r = \sqrt{100 + {(- 2)}^{2}}$

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

$r = \sqrt{100 + 4}$

Add $100$ and $4$ .

$r = \sqrt{104}$

Rewrite $104$ as $2^{2} \cdot 26$ .

Factor $4$ out of $104$ .

$r = \sqrt{4 (26)}$

Rewrite $4$ as $2^{2}$ .

$r = \sqrt{2^{2} \cdot 26}$

Pull terms out from under the radical.

$r = 2 \sqrt{26}$

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, $r = 2 \sqrt{26}$ and the center point is $(- 3, 6)$ . The equation for the circle is ${(x - (- 3))}^{2} + {(y - (6))}^{2} = {(2 \sqrt{26})}^{2}$ .

${(x - (- 3))}^{2} + {(y - (6))}^{2} = {(2 \sqrt{26})}^{2}$

Step 3

The circle equation is ${(x + 3)}^{2} + {(y - 6)}^{2} = 104$ .

${(x + 3)}^{2} + {(y - 6)}^{2} = 104$

Step 4

Find the Circle Through (7,4) with Center (-3,6) (7,4) , (-3,6)

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