Find the equation of the circle with diameter with end points (-3,2) and (4,1) ?

To find the equation of the circle with the given diameter, we first need to find the center of the circle.

The center of the circle is the midpoint of the diameter, which can be found by averaging the x-coordinates and y-coordinates of the two endpoints:

Center = ((-3 + 4)/2, (2 + 1)/2)
Center = (1/2, 3/2)

Next, we need to find the radius of the circle, which is half the length of the diameter:

Radius = distance between the center and one of the endpoints
Radius = sqrt((1/2 - (-3))^2 + (3/2 - 2)^2)
Radius = sqrt((5/2)^2 + (1/2)^2)
Radius = sqrt(25/4 + 1/4)
Radius = sqrt(26)/2

Now that we have the center and radius of the circle, we can write the equation of the circle in standard form:

(x - h)^2 + (y - k)^2 = r^2
where (h, k) is the center of the circle and r is the radius.

Plugging in the values we found:

(x - 1/2)^2 + (y - 3/2)^2 = (sqrt(26)/2)^2
(x - 1/2)^2 + (y - 3/2)^2 = 26/4
(x - 1/2)^2 + (y - 3/2)^2 = 13/2

Therefore, the equation of the circle with diameter with end points (-3,2) and (4,1) is:

(x - 1/2)^2 + (y - 3/2)^2 = 13/2

The center of the circle is the midpoint of the diameter, which is $$\left ( \frac{-3+4}{2}, \frac{2+1}{2} \right ) = \left ( \frac{1}{2},\frac{3}{2} \right )$$

The radius of the circle is half the length of the diameter, which is $$\frac{1}{2} \sqrt{(-3-4)^2 + (2-1)^2} = \frac{5\sqrt{2}}{2}$$

Therefore, the equation of the circle is $$\left ( x - \frac{1}{2} \right )^2 + \left ( y - \frac{3}{2} \right )^2 = \frac{50}{2}$$