Question

write the standard form of the equation of the circle for which the endpoints of a diameter are (0,0) and (4,-6)

Answer 1

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

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$d=\sqrt{(-6-0)^2+(4-0)^2}$

$d=\sqrt{36+16}$

$d=\sqrt{52}$

$d=2\sqrt{13}$

Now, radius is half of the diameter.

$r=\dfrac{2\sqrt{13}}{2}$

$r=\sqrt{13}$

Center of the circle is the midpoint of the endpoints of a diameter.

$Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$Center=\left(\dfrac{0+4}{2},\dfrac{0+(-6)}{2}\right)$

$Center=\left(\dfrac{4}{2},\dfrac{-6}{2}\right)$

$Center=\left(2,-3\right)$

Standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center and r is radius.

The center of the circle is (2,-3) and radius is $\sqrt{13}$. So,

$(x-2)^2+(y-(-3))^2=(\sqrt{13})^2$

$(x-2)^2+(y+3)^2=13$

Therefore, the standard form of the circle is $(x-2)^2+(y+3)^2=13$.