Find the standard form for the equation of a circle \( (x-h)^{2}+(y-k)^{2}=r^{2} \) with a diameter that has endpoints \( (-3,6) \) and \( (5,-8) \). \( h=\square \) \( k=\square \) \( r=\square \)

To find the standard form of the equation of a circle using the endpoints of the diameter, we first need to find the center \((h, k)\) of the circle. The center is the midpoint of the diameter endpoints. Calculating the midpoint of the points \((-3, 6)\) and \((5, -8)\): \[ h = \frac{-3 + 5}{2} = \frac{2}{2} = 1 \] \[ k = \frac{6 + (-8)}{2} = \frac{-2}{2} = -1 \] Next, we calculate the radius \(r\). The radius is half the distance between the two endpoints of the diameter. The distance \(d\) is given by: \[ d = \sqrt{(5 - (-3))^2 + (-8 - 6)^2} = \sqrt{(5 + 3)^2 + (-8 - 6)^2} = \sqrt{8^2 + (-14)^2} = \sqrt{64 + 196} = \sqrt{260} = 2\sqrt{65} \] Thus, the radius \(r\) will be: \[ r = \frac{d}{2} = \frac{2\sqrt{65}}{2} = \sqrt{65} \] Now we can plug these values into the standard form of the circle's equation: \((x - h)^{2} + (y - k)^{2} = r^{2}\) This results in: \[ (x - 1)^{2} + (y + 1)^{2} = (\sqrt{65})^{2} \] Thus, we have: \[ (x - 1)^{2} + (y + 1)^{2} = 65 \] So, the values are: \(h=1\) \(k=-1\) \(r=\sqrt{65}\)