44.0k views
5 votes
Write an equation for the circle with endpoints of a diameter (9, 4) and (-3, -2)

User Shawnest
by
8.9k points

1 Answer

4 votes
if the diameter is at (9,4) and (-3,-2), then the center of the circle is the midpoint of that segment.

and since we know that the radius of a circle is half of the diameter, whatever long that diameter segment is, the radius is half that.


\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 9 &,& 4~) % (c,d) &&(~ -3 &,& -2~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-3+9}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \stackrel{center}{(3~,~1)}


\bf -------------------------------\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ 9 &,& 4~) % (c,d) &&(~ -3 &,& -2~) \end{array} \\\\\\ d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((-3-9)^2+(-2-4)^2)\implies d=√((-12)^2+(-6)^2) \\\\\\ d=√(180)\qquad \qquad \qquad radius=\cfrac{√(180)}{2}


\bf -------------------------------\\\\ \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{3}{ h},\stackrel{1}{ k})\qquad \qquad radius=\stackrel{(√(180))/(2)}{ r} \\\\\\ (x-3)^2+(y-1)^2=\left( (√(180))/(2) \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(√(180))^2}{2^2} \\\\\\ (x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45
User NonlinearFruit
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories