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avatar+4221 

For what value of \(c\) will the circle with equation \(x^2 - 10x + y^2 + 6y + c = 0\) have a radius of length 1?

 Feb 5, 2018

Best Answer 

 #1
avatar+18328 
+1

X^2 -10x   + 25   +  y^2 + 6y   +   9   = -c   + 25 + 9

(x-5)^2  +  (y+3)^2  = -c  + 34           

 

-c+34 = 1

-c=-33

c = 33

 Feb 5, 2018
edited by Guest  Feb 5, 2018
edited by Guest  Feb 5, 2018
 #1
avatar+18328 
+1
Best Answer

X^2 -10x   + 25   +  y^2 + 6y   +   9   = -c   + 25 + 9

(x-5)^2  +  (y+3)^2  = -c  + 34           

 

-c+34 = 1

-c=-33

c = 33

ElectricPavlov Feb 5, 2018
edited by Guest  Feb 5, 2018
edited by Guest  Feb 5, 2018
 #2
avatar+4221 
+1

Thanks, EP! Amazing!

 Feb 6, 2018
 #3
avatar+183 
+2

Completing the square gives us \((x - 5)^2 + (y + 3)^2 = 34 - c\) . Since we want the radius to be 1, we must have \(34 - c = 1^2\) . It follows that \(c = \boxed{33}\).

 Feb 6, 2018

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