Answer:
[tex](x-3)^2+(y+4)^2=25[/tex]
Step-by-step explanation:
We start by writing the general form of the equation of a circle of radius R centered at [tex](x_0,y_0)[/tex]and creating three equations (one for each unknown: [tex]x_0[/tex], [tex]y_0[/tex], and the radius R):
[tex](x-x_0)^2+(y-y_0)^2=R^2[/tex]
1) If the circle passes through (0,0) then we should have that the equation above holds true:
[tex](x-x_0)^2+(y-y_0)^2=R^2\\(0-x_0)^2+(y-y_0)^2=R^2\\x_0^2+y_0^2=R^2\\R^2-x_0^2-y_0^2=0[/tex]
2) If the circle passes through (6,0) then we should have that the equation above holds true, and we also can use the important result from part 1) ([tex]R^2-x_0^2-y_0^2=0[/tex]) to solve for [tex]x_0[/tex]:
[tex](x-x_0)^2+(y-y_0)^2=R^2\\(6-x_0)^2+(y-y_0)^2=R^2\\36-12x_0+x_0^2+y_0^2=R^2\\-12x_0+36=R^2-x_0^2-y_0^2=0\\36=12x_0\\x_0=\frac{36}{12} =3[/tex]
3) If the circle passes through (0,-8) then we should have that the general equation of the circle above holds true, and we also can use the important result from part 1) ([tex]R^2-x_0^2-y_0^2=0[/tex]) to solve for [tex]y_0[/tex]:
[tex](x-x_0)^2+(y-y_0)^2=R^2\\(0-x_0)^2+(-8-y_0)^2=R^2\\x_0^2+64+16y_0+y_0^2=R^2\\64+16y_0=R^2-x_0^2-y_o^2=0\\16y_0=-64\\y_0=-\frac{64}{16}= -4[/tex]
4) and finally, we use the results for [tex]x_0[/tex] and [tex]y_0[/tex] of parts 2) and 3) back into the equation in part 1) to solve for [tex]R^2[/tex]:
[tex]R^2-x_0^2-y_0^2=0\\R^2-(3)^2-(-4)^2=0\\R^2-9-16=0\\\R^2=9+16=25\\[/tex]
Then, replacing [tex]x_0[/tex], [tex]y_0[/tex], and [tex]R^2[/tex] for the values we found, the equation of the circle becomes:
[tex](x-x_0)^2+(y-y_0)^2=R^2\\(x-3)^2+(y+4)^2=25[/tex]
