Examples

(3, 4)

$(3, 4)$ ,

(2, 5)

$(2, 5)$

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

(3, 4)

$(3, 4)$ and

(2, 5)

$(2, 5)$ .

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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{{(2 - 3)}^{2} + {(5 - 4)}^{2}}

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

Step 1.3

Simplify.

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Step 1.3.1

Subtract

3

$3$ from

2

$2$ .

r = \sqrt{{(- 1)}^{2} + {(5 - 4)}^{2}}

$r = \sqrt{{(- 1)}^{2} + {(5 - 4)}^{2}}$

Step 1.3.2

Raise

- 1

$- 1$ to the power of

2

$2$ .

r = \sqrt{1 + {(5 - 4)}^{2}}

$r = \sqrt{1 + {(5 - 4)}^{2}}$

Step 1.3.3

Subtract

4

$4$ from

5

$5$ .

r = \sqrt{1 + 1^{2}}

$r = \sqrt{1 + 1^{2}}$

Step 1.3.4

One to any power is one.

r = \sqrt{1 + 1}

$r = \sqrt{1 + 1}$

Step 1.3.5

Add

1

$1$ and

1

$1$ .

r = \sqrt{2}

$r = \sqrt{2}$

r = \sqrt{2}

$r = \sqrt{2}$

r = \sqrt{2}

$r = \sqrt{2}$

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{2}

$r = \sqrt{2}$ and the center point is

(3, 4)

$(3, 4)$ . The equation for the circle is

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

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

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

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

Step 3

The circle equation is

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

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

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

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

Step 4

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