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, -6)) and ((6, -6)).
(h =)
(k =)
(r =)

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 \((-1,-6)\) and \((6,-6)\), \(h=\frac{-1 + 6}{2}=\frac{5}{2}\), \(k=\frac{-6+( - 6)}{2}=-6\).

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 \(d=\sqrt{(6-( - 1))^2+(-6-( - 6))^2}=\sqrt{(6 + 1)^2+0^2}=7\).
So \(r=\frac{7}{2}\).

Answer:

\(h=\frac{5}{2}\), \(k = - 6\), \(r=\frac{7}{2}\)