Answer:
[tex]\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}[/tex]
Step-by-step explanation:
The equation of a circle in standard form:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
(h, k) - center
r - radius
1. We have the equation:
[tex](x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2[/tex]
2. We have the center (-4, 3) and the radius r = 6. Substitute:
[tex](x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36[/tex]
3. We have the endpoints of the diameter: (-1, 6) and (5, -4).
Midpoint of diameter is a center of a circle.
The formula of a midpoint:
[tex]\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)[/tex]
Substitute:
[tex]h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1[/tex]
The center is in (2, 1).
The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.
The formula of a distance between two points:
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Substitute the coordinates of the points (2, 1) and (5, -4):
[tex]r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}[/tex]
Finally we have:
[tex](x-2)^2+(y-1)^2=(\sqrt{34})^2[/tex]
