Question

If the end points of the diameter of a circle are (8,6) and (2,0), what is the standard form equation of the circle

Answer 1

Answer:

$\large\boxed{(x-5)^2+(y-3)^2=5^2\to(x-5)^2+(y-3)^2=25}$

Step-by-step explanation:

The standard form of an equation of a circle:

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

(h, k) - center

r - radius

We have the end points of a diameter (8, 6) and (2, 0). The midpoint of a diameter is a center of a circle. Half of a length of diameter is a length of a radius.

The formula of a midpoint of a segment:

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

Substitute:

$x=\dfrac{8+2}{2}=\dfrac{10}{2}=5\\\\y=\dfrac{6+0}{2}=\dfrac{6}{2}=3$

The center of a circle is (5, 3).

The formula of a length of a segment:

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

Substitute:

$d=\sqrt{(2-8)^2+(0-6)^2}=\sqrt{(-8)^2-(-6)^2}=\sqrt{64+36}=\sqrt{100}=10$

The length of a diameter is 10 units. The length of a radius r = 10 : 2 = 5 units.

Finally we have the equation of a circle:

$(x-5)^2+(y-3)^2=5^2$