Question 117154
The general equation for a circle is {{{(x-h)^2+(y-k)^2=r^2}}} where (h,k) is the center and r is the radius. Since we know the center is (2,3), we know that the equation looks like


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



Now we only need the radius. Since the point A(5,7) is on the circle, we need to find the distance from A to the center to find the radius

Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(2,3\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(5,7\right)]


{{{d=sqrt((2-5)^2+(3-7)^2)}}} Plug in {{{x[1]=2}}}, {{{x[2]=5}}}, {{{y[1]=3}}}, {{{y[2]=7}}}


{{{d=sqrt((-3)^2+(-4)^2)}}} Evaluate {{{2-5}}} to get -3. Evaluate {{{3-7}}} to get -4. 


{{{d=sqrt(9+16)}}} Square each value


{{{d=sqrt(25)}}} Add


{{{d=5}}} Simplify the square root  (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)


So the distance between (2,3) and (5,7) is 5 units. This means the radius is 5 units.


So {{{r=5}}} which means {{{r^2=5^2=25}}}



So the equation of the circle is {{{(x-2)^2+(y-3)^2=25}}}