Question 737637
Write the equation of the circle x^2+10x+y^2-6y+29=0 in standard form. Find the radius and center of the circle. 
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{{{x^2+10x+y^2-6y+29=0}}}
{{{x^2+10x + y^2-6y = -29}}}
Complete the squares for x and y separately.
{{{x^2+10x}}}
In the form {{{ax^2 + bx + c}}}, a = 1 & b = 10
Add c, c = (b/2a)^2
c = (10/2)^2 = 25
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y^2-6y
c = (-6/2)^2 = 9
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{{{x^2+10x25 + y^2-6y+9 = -29 + 25 + 9}}}
{{{x^2+10x25 + y^2-6y+9 = 5}}}
{{{(x+5)^2 + (y-3)^2 = 5}}} is "Standard Form"
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5 = r^2 --> {{{radius = sqrt(5)}}}
The center is (-5,3)