Question 388029
Recall the equation of a circle:

{{{(x-h)^2+(y-k)^2 = r^2 }}}where (h,k) is the circle's center and {{{r }}}is the radius.

In this problem, we need to complete the square to get the equation in the above form.



{{{x^2 + y^2 - 10x + 4y + 4 = 0}}}.......you just need to realize that you take half of the linear term (x and y)

{{{(x-5)^2 - 25+(y+2)^2 - 4 = -4 }}}

{{{(x-5)^2 +(y+2)^2 = -4 + 25 + 4}}}

{{{(x-5)^2 +(y+6)^2 =25}}}



{{{(x-5)^2 +(y+6)^2 = 5^2}}}...............{{{r^2=5^2}}}...>....{{{r=5}}}

It's obvious from above that the center of the circle is (5,-2) and radius is {{{5}}}.



*[invoke Plot_a_circle 5, -2, 5]