Precalculus Examples

(1, - 2)

$(1, - 2)$ ,

(3, 6)

$(3, 6)$

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

(1, - 2)

$(1, - 2)$ and

(3, 6)

$(3, 6)$ .

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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}^{2} + {(y_{2} - y_{1})}_{2}^{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 - 1)}^{2} + {(6 - (- 2))}^{2}}

$r = \sqrt{{(3 - 1)}^{2} + {(6 - (- 2))}^{2}}$

Step 1.3

Simplify.

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Step 1.3.1

Subtract

1

$1$ from

3

$3$ .

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

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

Step 1.3.2

Raise

2

$2$ to the power of

2

$2$ .

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

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

Step 1.3.3

Multiply

- 1

$- 1$ by

- 2

$- 2$ .

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

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

Step 1.3.4

Add

6

$6$ and

2

$2$ .

r = \sqrt{4 + 8^{2}}

$r = \sqrt{4 + 8^{2}}$

Step 1.3.5

Raise

8

$8$ to the power of

2

$2$ .

r = \sqrt{4 + 64}

$r = \sqrt{4 + 64}$

Step 1.3.6

Add

4

$4$ and

64

$64$ .

r = \sqrt{68}

$r = \sqrt{68}$

Step 1.3.7

Rewrite

68

$68$ as

2^{2} \cdot 17

$2^{2} \cdot 17$ .

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Step 1.3.7.1

Factor

4

$4$ out of

68

$68$ .

r = \sqrt{4 (17)}

$r = \sqrt{4 (17)}$

Step 1.3.7.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

r = \sqrt{2^{2} \cdot 17}

$r = \sqrt{2^{2} \cdot 17}$

r = \sqrt{2^{2} \cdot 17}

$r = \sqrt{2^{2} \cdot 17}$

Step 1.3.8

Pull terms out from under the radical.

r = 2 \sqrt{17}

$r = 2 \sqrt{17}$

r = 2 \sqrt{17}

$r = 2 \sqrt{17}$

r = 2 \sqrt{17}

$r = 2 \sqrt{17}$

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 = 2 \sqrt{17}

$r = 2 \sqrt{17}$ and the center point is

(1, - 2)

$(1, - 2)$ . The equation for the circle is

{(x - (1))}^{2} + {(y - (- 2))}^{2} = {(2 \sqrt{17})}^{2}

${(x - (1))}^{2} + {(y - (- 2))}^{2} = {(2 \sqrt{17})}^{2}$ .

{(x - (1))}^{2} + {(y - (- 2))}^{2} = {(2 \sqrt{17})}^{2}

${(x - (1))}^{2} + {(y - (- 2))}^{2} = {(2 \sqrt{17})}^{2}$

Step 3

The circle equation is

{(x - 1)}^{2} + {(y + 2)}^{2} = 68

${(x - 1)}^{2} + {(y + 2)}^{2} = 68$ .

${(x - 1)}^{2} + {(y + 2)}^{2} = 68$

Step 4

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