Given:
The endpoints of a diameter are (0,0) and (4,-6).
To find:
The equation of the circle.
Solution:
The endpoints of a diameter are (0,0) and (4,-6). So, the length of the diameter is
[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]d=\sqrt{(-6-0)^2+(4-0)^2}[/tex]
[tex]d=\sqrt{36+16}[/tex]
[tex]d=\sqrt{52}[/tex]
[tex]d=2\sqrt{13}[/tex]
Now, radius is half of the diameter.
[tex]r=\dfrac{2\sqrt{13}}{2}[/tex]
[tex]r=\sqrt{13}[/tex]
Center of the circle is the midpoint of the endpoints of a diameter.
[tex]Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]
[tex]Center=\left(\dfrac{0+4}{2},\dfrac{0+(-6)}{2}\right)[/tex]
[tex]Center=\left(\dfrac{4}{2},\dfrac{-6}{2}\right)[/tex]
[tex]Center=\left(2,-3\right)[/tex]
Standard form of a circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
where, (h,k) is center and r is radius.
The center of the circle is (2,-3) and radius is [tex]\sqrt{13}[/tex]. So,
[tex](x-2)^2+(y-(-3))^2=(\sqrt{13})^2[/tex]
[tex](x-2)^2+(y+3)^2=13[/tex]
Therefore, the standard form of the circle is [tex](x-2)^2+(y+3)^2=13[/tex].
