Find the Circle Through (-3,5) with Center (-1,-5) (-1,-5) , (-3,5)

$(- 1, - 5)$ , $(- 3, 5)$

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 $(- 1, - 5)$ and $(- 3, 5)$ .

Use the distance formula to determine the distance between the two points.

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

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

$r = \sqrt{{((- 3) - (- 1))}^{2} + {(5 - (- 5))}^{2}}$

Simplify.

Multiply $- 1$ by $- 1$ .

$r = \sqrt{{(- 3 + 1)}^{2} + {(5 - (- 5))}^{2}}$

Add $- 3$ and $1$ .

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

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

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

Multiply $- 1$ by $- 5$ .

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

Add $5$ and $5$ .

$r = \sqrt{4 + 10^{2}}$

Raise $10$ to the power of $2$ .

$r = \sqrt{4 + 100}$

Add $4$ and $100$ .

$r = \sqrt{104}$

Rewrite $104$ as $2^{2} \cdot 26$ .

Factor $4$ out of $104$ .

$r = \sqrt{4 (26)}$

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

$r = \sqrt{2^{2} \cdot 26}$

Pull terms out from under the radical.

$r = 2 \sqrt{26}$

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, $r = 2 \sqrt{26}$ and the center point is $(- 1, - 5)$ . The equation for the circle is ${(x - (- 1))}^{2} + {(y - (- 5))}^{2} = {(2 \sqrt{26})}^{2}$ .

${(x - (- 1))}^{2} + {(y - (- 5))}^{2} = {(2 \sqrt{26})}^{2}$

Step 3

The circle equation is ${(x + 1)}^{2} + {(y + 5)}^{2} = 104$ .

${(x + 1)}^{2} + {(y + 5)}^{2} = 104$

Step 4

Find the Circle Through (-3,5) with Center (-1,-5) (-1,-5) , (-3,5)

Related Questions

Support LCMGCF.com by disabling your adblocker.