# Find the center of the radius of x^2+y^2-10x+4y+13=0.

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### 1 Answer

We'll have to write the equation of the circle in the standard form.

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

We'll have to complete the square in the given general form of the equation of the circle:

x^2 - 10x + c^2

-10x = -2*x*c

We'll divide by -2x:

c = 5

We'll raise to square:

c^2 = 25

x^2 - 10x + c^2 = x^2 - 10x + 25

y^2 + 4y + d^2

4y = 2*y*d

We'll divide by 2y:

d = 2

d^2 = 4

y^2 + 4y + d^2 = y^2 + 4y + 4

So, we'll add and subtract 25 and 4 to the general form of the circle equation:

x^2 - 10x + 25 + y^2 + 4y + 4 - 25 - 4 + 13 = 0

(x - 5)^2 + (y + 2)^2 - 16 = 0

**(x - 5)^2 + (y + 2)^2 = 16**

**Comparing the standard form of the circle equation and the resulted form, we'll obtain the coordinates of the center of the circle C(5 ; -2) and the radius of the circle, r = 4.**