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 $(-8,0)$ and $(8,-10)$.
$h=$
$k=$
$r=$
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Explanation:

Step1: Find the center \((h,k)\)

The center of the circle is the mid - point of the diameter. The mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
For the points \((-8,0)\) and \((8,-10)\), we have \(h=\frac{-8 + 8}{2}=0\) and \(k=\frac{0+( - 10)}{2}=-5\).

Step2: Find the radius \(r\)

The radius is half of the length of the diameter. The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
The length of the diameter \(d=\sqrt{(8-( - 8))^2+(-10 - 0)^2}=\sqrt{(16)^2+( - 10)^2}=\sqrt{256 + 100}=\sqrt{356}=2\sqrt{89}\).
So the radius \(r=\frac{d}{2}=\sqrt{89}\).

Answer:

\(h = 0\), \(k=-5\), \(r=\sqrt{89}\)