Question

Write the equation in stands form for the circle with center (0, -8) passing through (0, 3/2)

Answer 1

well, we know the circle passes through (0 , 3/2) so the distance from the center and that point is its radius, so

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{(3)/(2)})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ \stackrel{radius}{r}= \sqrt{(~~ 0- 0 ~~)^2 + (~~ (3)/(2)- (-8) ~~)^2} \implies r= \sqrt{0 + (~~ (3)/(2) +8 ~~)^2} \\\\\\ r=\sqrt{\left( (19)/(2) \right)^2}\implies r=\cfrac{19}{2} \\\\[-0.35em] ~\dotfill`

`\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{0}{h}~~,~~\underset{-8}{k})}\qquad \stackrel{radius}{\underset{(19)/(2)}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - 0 ~~ )^2 ~~ + ~~ ( ~~ y-(-8) ~~ )^2~~ = ~~\left( \cfrac{19}{2} \right)^2\implies x^2+(y+8)^2~~ = ~~\cfrac{361}{4}`