Find the Circle Through (-1,2) with Center (3,4) (3,4) , (-1,2)

$(3, 4)$ , $(- 1, 2)$

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 $(3, 4)$ and $(- 1, 2)$ .

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.

$r = \sqrt{{((- 1) - 3)}^{2} + {(2 - 4)}^{2}}$

Simplify.

Subtract $3$ from $- 1$ .

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

Raise $- 4$ to the power of $2$ .

$r = \sqrt{16 + {(2 - 4)}^{2}}$

Subtract $4$ from $2$ .

$r = \sqrt{16 + {(- 2)}^{2}}$

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

$r = \sqrt{16 + 4}$

Add $16$ and $4$ .

$r = \sqrt{20}$

Rewrite $20$ as $2^{2} \cdot 5$ .

Factor $4$ out of $20$ .

$r = \sqrt{4 (5)}$

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

$r = \sqrt{2^{2} \cdot 5}$

Pull terms out from under the radical.

$r = 2 \sqrt{5}$

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, $r = 2 \sqrt{5}$ and the center point is $(3, 4)$ . The equation for the circle is ${(x - (3))}^{2} + {(y - (4))}^{2} = {(2 \sqrt{5})}^{2}$ .

${(x - (3))}^{2} + {(y - (4))}^{2} = {(2 \sqrt{5})}^{2}$

Step 3

The circle equation is ${(x - 3)}^{2} + {(y - 4)}^{2} = 20$ .

${(x - 3)}^{2} + {(y - 4)}^{2} = 20$

Step 4

Find the Circle Through (-1,2) with Center (3,4) (3,4) , (-1,2)

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