Question 192952
Step 1: Find the center of the circle



To find the center of the circle, you need to find the midpoint of the diameter with (4,5) and (6,1) as the endpoints. So simply average the corresponding coordinates to get:


{{{x[mid]=(4+6)/2=10/2=5}}}. So the x-coordinate of the center is {{{x=5}}}


{{{y[mid]=(1+5)/2=6/2=3}}}. So the y-coordinate of the center is {{{y=3}}}



So the center is (5,3). Since the center of 


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


is (h,k), this means that {{{h=5}}} and {{{k=3}}}



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Step 2: Finding the radius



To find the radius, we need to find length of the diameter. To find the diameter's length, we need to find the distance from (4,5) to (6,1). To do that, we can use the distance formula:


{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}}


Note: The point *[Tex \LARGE \left(x_1,y_1\right)] is (4,5) which means that {{{x[1]=4}}} and {{{y[1]=5}}}. Similarly, the point *[Tex \LARGE \left(x_1,y_1\right)] is (6,1) which means that {{{x[2]=6}}} and {{{y[2]=1}}}.




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



{{{d=sqrt((4-6)^2+(5-1)^2)}}} Plug in {{{x[1]=4}}},  {{{x[2]=6}}}, {{{y[1]=5}}}, and {{{y[2]=1}}}.



{{{d=sqrt((-2)^2+(5-1)^2)}}} Subtract {{{6}}} from {{{4}}} to get {{{-2}}}.



{{{d=sqrt((-2)^2+(4)^2)}}} Subtract {{{1}}} from {{{5}}} 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}}}.



{{{d=sqrt(4*5)}}} Factor {{{20}}} to get {{{4*5}}}.



{{{d=sqrt(4)*sqrt(5)}}} Break up the square root.



{{{d=2*sqrt(5)}}} Take the square root of {{{4}}} to get {{{2}}}.



So the diameter is {{{2*sqrt(5)}}} units long. Take half of this length to get the radius:



{{{r=(2*sqrt(5))/2=(cross(2)*sqrt(5))/cross(2)=sqrt(5)}}}



So {{{r=sqrt(5)}}} which means that the radius is {{{sqrt(5)}}} units long



Since {{{r=sqrt(5)}}}, this means that {{{r^2=(sqrt(5))^2=5}}} or just {{{r^2=5}}}



{{{(x-h)^2+(y-k)^2=r^2}}} Start with the general formula for a circle



{{{(x-5)^2+(y-3)^2=5}}} Plug in {{{h=5}}} and {{{k=3}}} and {{{r^2=5}}}



So the equation of the circle with the points (4,5) and (6,1) as the endpoints of the diameter is {{{(x-5)^2+(y-3)^2=5}}}