Answer:
See below for answers and explanations
Step-by-step explanation:
Problem A
The equation of a circle is [tex](x-h)^2+(y-k)^2=r^2[/tex] where [tex](h,k)[/tex] is the center and [tex]r[/tex] is the radius, thus, we need to find [tex]r^2[/tex] using our center and given point, through which we can find our equation:
[tex](x-h)^2+(y-k)^2=r^2\\\\(0-3)^2+(-2-6)^2=r^2\\\\(3)^2+(-8)^2=r^2\\\\9+64=r^2\\\\73=r^2[/tex]
This means that the correct equation is [tex](x-3)^2+(y-6)^2=73[/tex]
Problem B
[tex](x-h)^2+(y-k)^2=r^2\\\\(x-(-4))^2+(y-(-2))^2=9^2\\\\(x+4)^2+(y+2)^2=81[/tex]
Thus, the correct equation is [tex](x+4)^2+(y+2)^2=81[/tex]
Problem C
Use the distance formula where [tex](x_1,y_1)\rightarrow(2,5)[/tex] and [tex](x_2,y_2)\rightarrow(-10,7)[/tex] to find the diameter of the circle with the given endpoints:
[tex]d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\\\d=\sqrt{(7-5)^2+(-10-2)^2}\\\\d=\sqrt{(2)^2+(-12)^2}\\\\d=\sqrt{4+144}\\\\d=\sqrt{148}\\\\d=2\sqrt{37}[/tex]
Since the radius is half the diameter, then [tex]r=\sqrt{37}[/tex], making [tex]r^2=37[/tex].
The midpoint of the two endpoints will give us the center, so the center is [tex]\displaystyle \biggr(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\biggr)=\biggr(\frac{2+(-10)}{2},\frac{5+7}{2}\biggr)=\biggr(\frac{-8}{2},\frac{12}{2}\biggr)=(-4,6)[/tex]
Thus, the correct equation is [tex](x-h)^2+(y-k)^2=r^2\rightarrow(x-(-4))^2+(y-6)^2=37\rightarrow(x+4)^2+(y-6)^2=37[/tex]
