Question

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle? A) (x + 6)2 + (y + 10)2 = 20 B) (x − 6)2 + (y − 10)2 = 20 C) (x + 6)2 + (y + 10)2 = 2 5 D) (x − 6)2 + (y − 10)2 = 2 5

Answer 1

Answer:

$\large\boxed{A.\ (x+6)^2+(y+10)^2=20}$

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 endpoints of the diameter of a circle (-8, -6) and (-4, -14).

The midpoint of a diameter is a center of a circle.

The formula of a midpoint:

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

Substitute:

$x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10$

We have h = -6 and k = -10.

The radius is the distance between a center and the point on a circumference of a circle.

The formula of a distance between two points:

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

Substitute (-6, -10) and (-8, -6):

$r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}$

Finally we have

$(x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20$

Answer 2

Answer:

It's A...Just had it but i Chose B but it's A

Step-by-step explanation: