Question 288447
 For the given circle, find the x-intercepts and the y-intercepts. 
x^2+y^2-10x+4y+13=0 
This is what I have so far:
To find the x-intercepts let y=0
X^2-10x+13=0
this equation doesn't factor into integers as 13 is a prime number 
Most of life is not integers.
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*[invoke solve_quadratic_equation 1,-10,13]

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The x-intercepts are not integers.
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To get the y-intercepts:
y^2 + 4y + 13 = 0
*[invoke solve_quadratic_equation 1,4,13]
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There are no real numbers solutions --> no y-intercepts.
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If you complete the square:
x^2+y^2-10x+4y= -13
x^2-10x+25 + y^2+4y+4 = 16
{{{x-5)^2 + (y+2)^2 = 4^2}}}
The center of the circle is (5,-2) and its radius is 4.
It's 5 units to the right with a radius of 4, --> no y-intercepts.