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}{(15)/(2)}~,~\stackrel{y_2}{1})\qquad \qquadd = √(( x_2- x_1)^2 + ( y_2- y_1)^2)\\\\\\\stackrel{radius}{r}=\sqrt{[(15)/(2) - 0]^2 + [1 - (-3)]^2}\implies r=\sqrt{\left( (15)/(2) \right)^2 + (1+3)^2}\\\\\\r=\sqrt{\left( (15)/(2) \right)^2 +4^2}\implies r=\sqrt{(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{(17)/(2)}{ r} \\\\\\\ [x-0]^2~~ + ~~[y-(-3)]^2~~ = ~~\left( \cfrac{17}{2} \right)^2\implies x^2+(y+3)^2 = \cfrac{289}{4}`