Precalculus Examples

$(2, 7)$ , $(3, 4)$

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 $(3, 4)$ .

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Step 1.1

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Step 1.2

Substitute the actual values of the points into the distance formula.

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

Step 1.3

Simplify.

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Step 1.3.1

Subtract $2$ from $3$ .

$r = \sqrt{1^{2} + {(4 - 7)}^{2}}$

Step 1.3.2

One to any power is one.

$r = \sqrt{1 + {(4 - 7)}^{2}}$

Step 1.3.3

Subtract $7$ from $4$ .

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

Step 1.3.4

Raise $- 3$ to the power of $2$ .

$r = \sqrt{1 + 9}$

Step 1.3.5

Add $1$ and $9$ .

$r = \sqrt{10}$

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, $r = \sqrt{10}$ and the center point is $(2, 7)$ . The equation for the circle is ${(x - (2))}^{2} + {(y - (7))}^{2} = {(\sqrt{10})}^{2}$ .

${(x - (2))}^{2} + {(y - (7))}^{2} = {(\sqrt{10})}^{2}$

Step 3

The circle equation is ${(x - 2)}^{2} + {(y - 7)}^{2} = 10$ .

${(x - 2)}^{2} + {(y - 7)}^{2} = 10$

Step 4

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