To find the equation of the circle that has the segment PQ as a diameter, we need to determine the center and the radius of the circle.
### Step 1: Find the Midpoint of PQ (Center of the Circle)
The center of the circle is the midpoint of the segment PQ. The formula for the midpoint [tex]\( M \)[/tex] of a segment with endpoints [tex]\( P(x_1, y_1) \)[/tex] and [tex]\( Q(x_2, y_2) \)[/tex] is:
[tex]\[
M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\][/tex]
For [tex]\( P = (3, 1) \)[/tex] and [tex]\( Q = (-3, -7) \)[/tex]:
[tex]\[
\text{Center}_x = \frac{3 + (-3)}{2} = \frac{0}{2} = 0
\][/tex]
[tex]\[
\text{Center}_y = \frac{1 + (-7)}{2} = \frac{-6}{2} = -3
\][/tex]
So, the center of the circle is [tex]\( (0, -3) \)[/tex].
### Step 2: Find the Radius of the Circle
The radius is half the distance of the segment PQ. The distance [tex]\( d \)[/tex] between points [tex]\( P(x_1, y_1) \)[/tex] and [tex]\( Q(x_2, y_2) \)[/tex] is given by:
[tex]\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\][/tex]
Then, the radius [tex]\( r \)[/tex] is [tex]\( \frac{d}{2} \)[/tex].
[tex]\[
d = \sqrt{(-3 - 3)^2 + (-7 - 1)^2} = \sqrt{(-6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10
\][/tex]
The radius is:
[tex]\[
r = \frac{10}{2} = 5
\][/tex]
### Step 3: Write the Equation of the Circle
The equation of a circle with center [tex]\( (h, k) \)[/tex] and radius [tex]\( r \)[/tex] is:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Substituting [tex]\( h = 0 \)[/tex], [tex]\( k = -3 \)[/tex], and [tex]\( r = 5 \)[/tex]:
[tex]\[
(x - 0)^2 + (y + 3)^2 = 5^2
\][/tex]
[tex]\[
x^2 + (y + 3)^2 = 25
\][/tex]
Thus, the equation of the circle is:
[tex]\[
(x - 0)^2 + (y + 3)^2 = 25
\][/tex]
Therefore, the equation of the circle that has segment PQ as a diameter is:
[tex]\[
(x - 0)^2 + (y + 3)^2 = 25
\][/tex]
