Algebra Examples

(2, 7)

$(2, 7)$ ,

(3, 4)

$(3, 4)$

Step 1

Find the radius

r

$r$ 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

$r$ is the distance between

(2, 7)

$(2, 7)$ and

(3, 4)

$(3, 4)$ .

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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}}

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

Step 1.2

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

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

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

Step 1.3

Simplify.

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Step 1.3.1

Subtract

2

$2$ from

3

$3$ .

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

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

Step 1.3.2

One to any power is one.

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

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

Step 1.3.3

Subtract

7

$7$ from

4

$4$ .

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

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

Step 1.3.4

Raise

- 3

$- 3$ to the power of

2

$2$ .

r = \sqrt{1 + 9}

$r = \sqrt{1 + 9}$

Step 1.3.5

Add

1

$1$ and

9

$9$ .

r = \sqrt{10}

$r = \sqrt{10}$

r = \sqrt{10}

$r = \sqrt{10}$

r = \sqrt{10}

$r = \sqrt{10}$

Step 2

{(x - h)}^{2} + {(y - k)}^{2} = r^{2}

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the equation form for a circle with

r

$r$ radius and

(h, k)

$(h, k)$ as the center point. In this case,

r = \sqrt{10}

$r = \sqrt{10}$ and the center point is

(2, 7)

$(2, 7)$ . The equation for the circle is

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

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

{(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

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

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

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

Step 4

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