SOLUTION:
Case: Circle equations
Method:
The endpoints of the diameter are P(-2, -1) and Q(-6, -3)
The diameter therefore:
[tex]\begin{gathered} d^2=(-6--2)^2+(-3--1)^2 \\ d^2=(-6+2)^2+(-3+1)^2 \\ d^2=(-4)^2+(-2)^2 \\ d^2=16+4 \\ d^2=20 \\ d=\sqrt{20} \\ d=\sqrt{4\times5} \\ d=2\sqrt{5} \end{gathered}[/tex]
The radius therefore is:
[tex]\begin{gathered} diameter=2\sqrt{5} \\ radius \\ r=\frac{2\sqrt{5}}{2} \\ r=\sqrt{5} \end{gathered}[/tex]
The equation of a circle is given as:
[tex]\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-a)^2+(y-b)^2=(\sqrt{5})^2 \\ (x-a)^2+(y-b)^2=5 \end{gathered}[/tex]
The center (a, b) is the midpoint between P and Q
[tex]\begin{gathered} x_m=\frac{x_1+x_2}{2} \\ x_m=\frac{-2-6}{2} \\ x_m=\frac{-8}{2} \\ x_m=-4 \\ AND \\ y_m=\frac{y_1+y_2}{2} \\ y_m=\frac{-1-3}{2} \\ y_m=\frac{-4}{2} \\ y_m=-2 \\ The\text{ }center \\ (-4,-2) \end{gathered}[/tex]
The center (a,b) is (-4, -2)
Final answer:
[tex](x-[-4])^2+(y-[-2])^2=5[/tex]
