Precalculus Examples

$(- 3, 8)$ , $(7, 6)$

Step 1

The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circumference of the circle. The given end points of the diameter are $(- 3, 8)$ and $(7, 6)$ . The center point of the circle is the center of the diameter, which is the midpoint between $(- 3, 8)$ and $(7, 6)$ . In this case the midpoint is $(2, 7)$ .

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Step 1.1

Use the midpoint formula to find the midpoint of the line segment.

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Step 1.2

Substitute in the values for $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ .

$(\frac{- 3 + 7}{2}, \frac{8 + 6}{2})$

Step 1.3

Add $- 3$ and $7$ .

$(\frac{4}{2}, \frac{8 + 6}{2})$

Step 1.4

Divide $4$ by $2$ .

$(2, \frac{8 + 6}{2})$

Step 1.5

Cancel the common factor of $8 + 6$ and $2$ .

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Step 1.5.1

Factor $2$ out of $8$ .

$(2, \frac{2 \cdot 4 + 6}{2})$

Step 1.5.2

Factor $2$ out of $6$ .

$(2, \frac{2 \cdot 4 + 2 \cdot 3}{2})$

Step 1.5.3

Factor $2$ out of $2 \cdot 4 + 2 \cdot 3$ .

$(2, \frac{2 \cdot (4 + 3)}{2})$

Step 1.5.4

Cancel the common factors.

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Step 1.5.4.1

Factor $2$ out of $2$ .

$(2, \frac{2 \cdot (4 + 3)}{2 (1)})$

Step 1.5.4.2

Cancel the common factor.

$(2, \frac{2 \cdot (4 + 3)}{2 \cdot 1})$

Step 1.5.4.3

Rewrite the expression.

$(2, \frac{4 + 3}{1})$

Step 1.5.4.4

Divide $4 + 3$ by $1$ .

$(2, 4 + 3)$

Step 1.6

Add $4$ and $3$ .

$(2, 7)$

Step 2

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 $(- 3, 8)$ .

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Step 2.1

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 2.2

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

$r = \sqrt{{((- 3) - 2)}^{2} + {(8 - 7)}^{2}}$

Step 2.3

Simplify.

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Step 2.3.1

Subtract $2$ from $- 3$ .

$r = \sqrt{{(- 5)}^{2} + {(8 - 7)}^{2}}$

Step 2.3.2

Raise $- 5$ to the power of $2$ .

$r = \sqrt{25 + {(8 - 7)}^{2}}$

Step 2.3.3

Subtract $7$ from $8$ .

$r = \sqrt{25 + 1^{2}}$

Step 2.3.4

One to any power is one.

$r = \sqrt{25 + 1}$

Step 2.3.5

Add $25$ and $1$ .

$r = \sqrt{26}$

Step 3

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ is the equation form for a circle with $r$ radius and $(h, k)$ as the center point. In this case, $r = \sqrt{26}$ and the center point is $(2, 7)$ . The equation for the circle is ${(x - (2))}^{2} + {(y - (7))}^{2} = {(\sqrt{26})}^{2}$ .

${(x - (2))}^{2} + {(y - (7))}^{2} = {(\sqrt{26})}^{2}$

Step 4

The circle equation is ${(x - 2)}^{2} + {(y - 7)}^{2} = 26$ .

${(x - 2)}^{2} + {(y - 7)}^{2} = 26$

Step 5