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For what value of $c$ will the circle with equation $x^2 + 8x + y^2 + 4y + c = 0$ have a radius of length 3?

Guest Jan 14, 2018

Best Answer 

 #1
avatar+13635 
+2

Equation of a circle :  (x-h)^2  +  (y-k)^2  = r ^2        h, k being the center and r = radius

 

SO let's get to that form

x^2 +   8x       + y^2 + 4y    =  -c

Kinda 'complete the squares' now

(x+4)^2   + (y+2)^2  =  -c + 20            (if you complete the squares on the left side ...you will see we added 20)

 

so    -c + 20  = r^2 = 3^2 = 9     therefore   -c = -11    or  c= 11        center of circle (-4,-2)   radius = 3

ElectricPavlov  Jan 14, 2018
edited by ElectricPavlov  Jan 14, 2018
 #1
avatar+13635 
+2
Best Answer

Equation of a circle :  (x-h)^2  +  (y-k)^2  = r ^2        h, k being the center and r = radius

 

SO let's get to that form

x^2 +   8x       + y^2 + 4y    =  -c

Kinda 'complete the squares' now

(x+4)^2   + (y+2)^2  =  -c + 20            (if you complete the squares on the left side ...you will see we added 20)

 

so    -c + 20  = r^2 = 3^2 = 9     therefore   -c = -11    or  c= 11        center of circle (-4,-2)   radius = 3

ElectricPavlov  Jan 14, 2018
edited by ElectricPavlov  Jan 14, 2018
 #2
avatar+91097 
0

Nice, EP....!!!!

 

 

cool cool cool

CPhill  Jan 15, 2018

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