Question

Write the equation in standard form for the circle with center (0, -3) passing through (15/2 , 1)

Answer 1

Check the picture below, so the green line is really the radius of the circle, and we know its center.

$~~~~~~~~~~~~\textit{distance between 2 points}\\\\(\stackrel{x_1}{0}~,~\stackrel{y_1}{-3})\qquad(\stackrel{x_2}{\frac{15}{2}}~,~\stackrel{y_2}{1})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\\stackrel{radius}{r}=\sqrt{[\frac{15}{2} - 0]^2 + [1 - (-3)]^2}\implies r=\sqrt{\left( \frac{15}{2} \right)^2 + (1+3)^2}\\\\\\r=\sqrt{\left( \frac{15}{2} \right)^2 +4^2}\implies r=\sqrt{\frac{225}{4} + 16}\implies r=\sqrt{\cfrac{289}{4}}\implies r=\cfrac{17}{2}\\\\[-0.35em]\rule{34em}{0.25pt}$

$\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{0}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{\frac{17}{2}}{ r} \\\\\\\ [x-0]^2~~ + ~~[y-(-3)]^2~~ = ~~\left( \cfrac{17}{2} \right)^2\implies x^2+(y+3)^2 = \cfrac{289}{4}$