ANSWER
[tex](x-11)^2+(y+8)^2=74[/tex]
EXPLANATION
Given:
A circle with endpoints (18, -13) and (4, 3) as the diameter.
Desired Outcome:
Equation of the circle.
Determine the center of the circle using the midpoint formula
[tex](\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]
where:
x1 = 18
x2 = 4
y1 = -13
y2 = -3
Substitute the values:
[tex]\begin{gathered} (\frac{18+4}{2},\frac{-13-3}{2}) \\ =(\frac{22}{2},\frac{-16}{2}) \\ =(11,-8) \end{gathered}[/tex]
Therefore, the center (h, k) of the circle is (11, -8)
Determine the radius of the circle
[tex]r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}^[/tex]
where:
x1 = 11
x2 = 4
y1 = -8
y2 = -3
Substitute the values:
[tex]\begin{gathered} r=\sqrt{(4-11)^2+(-3--8)^2}^ \\ r=\sqrt{(-7)^2+(-5)^2} \\ r=\sqrt{49+25} \\ r=\sqrt{74} \end{gathered}[/tex]
Now, the equation of the circle is:
[tex]\begin{gathered} (x-h)^2+(y-k)^2=r^2 \\ (x-11)^2+(y--8)^2=(\sqrt{74})^2 \\ (x-11)^2+(y+8)^2=74 \end{gathered}[/tex]
Graph:
