Question

PLEASE HELP.

What is the equation of the circle in standard form where the center is (-16,-14) and another point on the circle is (-8,-8)

Answer 1

recall that the radius is the distance from the center of a circle to any point on the circle.

we know the center is at -16,-14, and we know that -8,-8 is a point on the circle, so the distance between both must be the radius.

`\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-16}~,~\stackrel{y_1}{-14})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{-8})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{radius}{r}=√([-8-(-16)]^2+[-8-(-14)]^2) \\\\\\ r=√((-8+16)^2+(-8+14)^2)\implies r=√(8^2+6^2) \\\\\\ r=√(100)\implies \boxed{r=10} \\\\[-0.35em] \rule{34em}{0.25pt}`

`\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-16}{ h},\stackrel{-14}{ k})\qquad \qquad radius=\stackrel{10}{ r} \\\\\\\ [x-(-16)]^2+[y-(-14)]^2=10^2\implies \blacktriangleright (x+16)^2+(y+14)^2=100 \blacktriangleleft`