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
Accepted Solution
A:
Answer:[tex]\large\boxed{A.\ (x+6)^2+(y+10)^2=20}[/tex]Step-by-step explanation:The standard form of an equation of a circle:[tex](x-h)^2+(y-k)^2=r^2[/tex](h, k) - centerr - radiusWe 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:[tex]\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)[/tex]Substitute:[tex]x=\dfrac{-8+(-4)}{2}=\dfrac{-12}{2}=-6\\\\y=\dfrac{-6+(-14)}{2}=\dfrac{-20}{2}=-10[/tex]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:[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Substitute (-6, -10) and (-8, -6):[tex]r=\sqrt{(-8-(-6))^2+(-6-(-10))^2}=\sqrt{(-2)^2+4^2}=\sqrt{4+16}=\sqrt{20}[/tex]Finally we have[tex](x-(-6))^2+(y-(-10))^2=(\sqrt{20})^2\\\\(x+6)^2+(y+10)^2=20[/tex]