Question

find the standard form for the equation of a circle $(x - h)^2+(y - k)^2=r^2$ with a diameter that has endpoints $(-1,-3)$ and $(7,-4)$. $h =$ $k =$ $r =$

Explanation:

Step1: Find the center (h, k)

The center of the circle is the midpoint of the diameter. The midpoint formula is $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$. For endpoints $(-1, -3)$ and $(7, -4)$, we have:
$h = \frac{-1 + 7}{2} = \frac{6}{2} = 3$
$k = \frac{-3 + (-4)}{2} = \frac{-7}{2} = -3.5$

Step2: Find the radius r

First, find the length of the diameter using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$d = \sqrt{(7 - (-1))^2 + (-4 - (-3))^2} = \sqrt{(8)^2 + (-1)^2} = \sqrt{64 + 1} = \sqrt{65}$
The radius is half of the diameter, so $r = \frac{\sqrt{65}}{2}$. Then $r^2 = (\frac{\sqrt{65}}{2})^2 = \frac{65}{4}$ (but we can also just find r as $\frac{\sqrt{65}}{2}$ for the value of r).

Answer:

$h = 3$
$k = -\frac{7}{2}$ (or -3.5)
$r = \frac{\sqrt{65}}{2}$