Find the Circle Through (3,0) with Center (3,-2) (3,0) , (3,-2)

$(3, 0)$ , $(3, - 2)$

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 $(3, - 2)$ and $(3, 0)$ .

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.

$r = \sqrt{{(3 - 3)}^{2} + {(0 - (- 2))}^{2}}$

Simplify.

Subtract $3$ from $3$ .

$r = \sqrt{0^{2} + {(0 - (- 2))}^{2}}$

Raising $0$ to any positive power yields $0$ .

$r = \sqrt{0 + {(0 - (- 2))}^{2}}$

Multiply $- 1$ by $- 2$ .

$r = \sqrt{0 + {(0 + 2)}^{2}}$

Add $0$ and $2$ .

$r = \sqrt{0 + 2^{2}}$

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

$r = \sqrt{0 + 4}$

Add $0$ and $4$ .

$r = \sqrt{4}$

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

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

Pull terms out from under the radical, assuming positive real numbers.

$r = 2$

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, $r = 2$ and the center point is $(3, - 2)$ . The equation for the circle is ${(x - (3))}^{2} + {(y - (- 2))}^{2} = {(2)}^{2}$ .

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

Step 3

The circle equation is ${(x - 3)}^{2} + {(y + 2)}^{2} = 4$ .

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

Step 4

Find the Circle Through (3,0) with Center (3,-2) (3,0) , (3,-2)

Related Questions

Support LCMGCF.com by disabling your adblocker.