Find the standard form for the equation of a circle \[ (x-h)^{2}+(y-k)^{2}=r^{2} \] with a diameter that has endpoints $(-2,9)$ and $(5,5)$. \[ \begin{array}{l} h= \\ k= \\ r= \end{array} \] Question Help: Video Submit Question Jump to Answer
Solution
Step 1 ：Given that the endpoints of the diameter are (-2,9) and (5,5), we can find the center of the circle by finding the midpoint of these two points. The midpoint formula is given by \(h = \frac{x_1 + x_2}{2}\) and \(k = \frac{y_1 + y_2}{2}\). So, \(h = \frac{-2 + 5}{2} = 1.5\) and \(k = \frac{9 + 5}{2} = 7.0\).
Step 2 ：The radius of the circle is half the length of the diameter. We can find the length of the diameter using the distance formula, which is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\), and then divide by 2 to get the radius. So, \(d = \sqrt{(5 - (-2))^2 + (5 - 9)^2} = 8.06225774829855\) and \(r = \frac{d}{2} = 4.031128874149275\).
Step 3 ：Finally, the standard form of the equation of the circle is \((x-h)^2 + (y-k)^2 = r^2\). Substituting the values of h, k, and r, we get \((x-1.5)^2 + (y-7.0)^2 = (4.031128874149275)^2\).
Step 4 ：\(\boxed{(x-1.5)^2 + (y-7.0)^2 = 16.25}\)
