Precalculus Examples

$(1, - 2)$ , $(3, 6)$

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 $(1, - 2)$ and $(3, 6)$ .

Tap for more steps...

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.

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

Step 1.3

Simplify.

Tap for more steps...

Step 1.3.1

Subtract $1$ from $3$ .

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

Step 1.3.2

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

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

Step 1.3.3

Multiply $- 1$ by $- 2$ .

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

Step 1.3.4

Add $6$ and $2$ .

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

Step 1.3.5

Raise $8$ to the power of $2$ .

$r = \sqrt{4 + 64}$

Step 1.3.6

Add $4$ and $64$ .

$r = \sqrt{68}$

Step 1.3.7

Rewrite $68$ as $2^{2} \cdot 17$ .

Tap for more steps...

Step 1.3.7.1

Factor $4$ out of $68$ .

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

Step 1.3.7.2

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

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

Step 1.3.8

Pull terms out from under the radical.

$r = 2 \sqrt{17}$

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, $r = 2 \sqrt{17}$ and the center point is $(1, - 2)$ . The equation for the circle is ${(x - (1))}^{2} + {(y - (- 2))}^{2} = {(2 \sqrt{17})}^{2}$ .

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

Step 3

The circle equation is ${(x - 1)}^{2} + {(y + 2)}^{2} = 68$ .

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

Step 4

Enter YOUR Problem