For what value of $c$ will the circle with equation $x^2 + 8x + y^2 + 4y + c = 0$ have a radius of length 3?
+36916 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
+36916 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
