Question 394404
A circle is in the form {{{(x-h)^2+(y-k)^2=r^2}}} where the center is (h,k) with radius 'r'



Since we know that the center is (3,4), we're given that {{{h=3}}} and {{{k=4}}}



Plug these values into the equation above to get {{{(x-3)^2+(y-4)^2=r^2}}}



So all we really need is the radius 'r'. We can find this by finding the distance from the center (3,4) to the point (7,9) (since this point lies on the circle)



Let's use the distance formula to find this distance



Note: *[Tex \LARGE \left(x_{1}, y_{1}\right)] is the first point *[Tex \LARGE \left(3,4\right)]. So this means that {{{x[1]=3}}} and {{{y[1]=4}}}.

Also, *[Tex \LARGE \left(x_{2}, y_{2}\right)] is the second point *[Tex \LARGE \left(7,9\right)].  So this means that {{{x[2]=7}}} and {{{y[2]=9}}}.



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



{{{d=sqrt((3-7)^2+(4-9)^2)}}} Plug in {{{x[1]=3}}},  {{{x[2]=7}}}, {{{y[1]=4}}}, and {{{y[2]=9}}}.



{{{d=sqrt((-4)^2+(4-9)^2)}}} Subtract {{{7}}} from {{{3}}} to get {{{-4}}}.



{{{d=sqrt((-4)^2+(-5)^2)}}} Subtract {{{9}}} from {{{4}}} to get {{{-5}}}.



{{{d=sqrt(16+(-5)^2)}}} Square {{{-4}}} to get {{{16}}}.



{{{d=sqrt(16+25)}}} Square {{{-5}}} to get {{{25}}}.



{{{d=sqrt(41)}}} Add {{{16}}} to {{{25}}} to get {{{41}}}.



Since the distance is {{{sqrt(41)}}} units, the radius is {{{r=sqrt(41)}}}. Square this value to get {{{r^2=(sqrt(41))=41}}}. So {{{r^2=41}}}



So the equation of the circle is {{{(x-3)^2+(y-4)^2=41}}}



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