Find the standard form of the equation of the circle with endpoints of a diameter at the points \( (7,6) \) and \( (-5,4) \). Type the standard form of the equation of this circle. \( \square \) (Type an equation.)

To find the standard form of the equation of the circle, we first need to determine the center and radius. The center \( (h, k) \) of the circle is the midpoint of the diameter points \( (7,6) \) and \( (-5,4) \). Calculating the midpoint: \[ h = \frac{7 + (-5)}{2} = \frac{2}{2} = 1 \] \[ k = \frac{6 + 4}{2} = \frac{10}{2} = 5 \] So, the center is \( (1, 5) \). Next, we find the radius \( r \) by calculating the distance from the center to one of the endpoints, say \( (7, 6) \): \[ r = \sqrt{(7 - 1)^2 + (6 - 5)^2} = \sqrt{(6)^2 + (1)^2} = \sqrt{36 + 1} = \sqrt{37} \] Now, the standard form of the equation of a circle is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] Substituting in our values of \( h, k, \) and \( r^2 \): \[ (x - 1)^2 + (y - 5)^2 = 37 \] So the standard form of the equation of the circle is: \[ (x - 1)^2 + (y - 5)^2 = 37 \]