Question

assignment 10.0: distance and midpoint formulas; circles
score: 60/100 answered: 8/12
question 9
find the standard form for the equation of a circle
$(x - h)^{2}+(y - k)^{2}=r^{2}$
with a diameter that has endpoints $(-6,-3)$ and $(5,-5)$.
$h=$
$k=$
$r=$
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Explanation:

Step1: Find the center of the circle (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 \((-6,-3)\) and \((5,-5)\), \(h=\frac{-6 + 5}{2}=-\frac{1}{2}\), \(k=\frac{-3+( - 5)}{2}=\frac{-3 - 5}{2}=-4\).

Step2: Find the radius (half of the length of the diameter)

The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For the points \((-6,-3)\) and \((5,-5)\), \(d=\sqrt{(5-( - 6))^2+(-5-( - 3))^2}=\sqrt{(5 + 6)^2+(-5 + 3)^2}=\sqrt{121 + 4}=\sqrt{125}=5\sqrt{5}\).
Then \(r=\frac{d}{2}=\frac{5\sqrt{5}}{2}\).

Answer:

\(h =-\frac{1}{2}\), \(k=-4\), \(r=\frac{5\sqrt{5}}{2}\)