Question 397031
You find the midpoint, which is the center of the circle. Since you found the midpoint to be (3,-2), this means that the center of the circle is (3,-2). This is only possible because the line segment PQ is the diameter.



Recall that the equation of the circle with radius 'r' and center (h,k) is {{{(x-h)^2+(y-k)^2=r^2}}}



So because the center of the circle is (3,-2), this means that {{{h=3}}} and {{{k=-2}}}. Plug these values into the equation above to get 



{{{(x-3)^2+(y-(-2))^2=r^2}}}



and simplify to get 



{{{(x-3)^2+(y+2)^2=r^2}}}



So all we need to find out is the value of {{{r^2}}}. So we need to find the value of the radius 'r'. It turns out that the distance from the center (3,-2) to either point will be equal to the radius (draw a picture if you're not sure)



So to find 'r', let's find the distance from P(5,-6) to the center (3,-2)



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(5,-6\right)]. So this means that {{{x[1]=5}}} and {{{y[1]=-6}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(3,-2\right)].  So this means that {{{x[2]=3}}} and {{{y[2]=-2}}}.



{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula.



{{{d=sqrt((5-3)^2+(-6--2)^2)}}} Plug in {{{x[1]=5}}},  {{{x[2]=3}}}, {{{y[1]=-6}}}, and {{{y[2]=-2}}}.



{{{d=sqrt((2)^2+(-6--2)^2)}}} Subtract {{{3}}} from {{{5}}} to get {{{2}}}.



{{{d=sqrt((2)^2+(-4)^2)}}} Subtract {{{-2}}} from {{{-6}}} to get {{{-4}}}.



{{{d=sqrt(4+(-4)^2)}}} Square {{{2}}} to get {{{4}}}.



{{{d=sqrt(4+16)}}} Square {{{-4}}} to get {{{16}}}.



{{{d=sqrt(20)}}} Add {{{4}}} to {{{16}}} to get {{{20}}}.



So this distance is {{{sqrt(20)}}} units. So {{{r=sqrt(20)}}}



This then means that {{{r^2=(sqrt(20))^2=20}}}



So {{{r^2=20}}}



Plug this last value into the equation from the beginning to get 



{{{(x-3)^2+(y+2)^2=20}}}




So the equation of the circle with a diameter whose endpoints are P(5,-6) and Q(1,2) is {{{(x-3)^2+(y+2)^2=20}}}



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Jim