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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?

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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

#1
+10761
+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
Sort:

#1
+10761
+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
#2
+80874
0

Nice, EP....!!!!

CPhill  Jan 15, 2018

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