Respuesta :
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.
[tex]\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)}[/tex]
[tex]\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 = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2} \\\\\\ d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}[/tex]
[tex]\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{\frac{\sqrt{180}}{2}}{ r} \\\\\\ (x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2} \\\\\\ (x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45[/tex]
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.
[tex]\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)}[/tex]
[tex]\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 = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{(-3-9)^2+(-2-4)^2}\implies d=\sqrt{(-12)^2+(-6)^2} \\\\\\ d=\sqrt{180}\qquad \qquad \qquad radius=\cfrac{\sqrt{180}}{2}[/tex]
[tex]\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{\frac{\sqrt{180}}{2}}{ r} \\\\\\ (x-3)^2+(y-1)^2=\left( \frac{\sqrt{180}}{2} \right)^2\implies (x-3)^2+(y-1)^2=\cfrac{(\sqrt{180})^2}{2^2} \\\\\\ (x-3)^2+(y-1)^2=\cfrac{180}{4}\implies (x-3)^2+(y-1)^2=45[/tex]
