+0

# what is the center of a circle and the radius of 4x^2+4x+4y^2-16y-127=0

+3
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what is the center of a circle and the radius of 4x^2+4x+4y^2-16y-127=0

Guest Jul 10, 2014

#2
+84238
+11

Alan has an interesting approach.......mine is a little different........but it should still work out the same...

Let's divide the original equation by 4....so we have

x^2 + x + y^2 - 4y - 127/4 =0          add  127/4 to both sides

x^2 + x + y^2 - 4y = 127/4                now, complete the sqaure on x and y

(x^2 + x + 1/4) + (y^2 - 4y + 4) =  127/4 + 1/4 + 4         simplifying, we have

(x + 1/2)^2  + (y - 2)^2  = 144/4

(x + 1/2)^2 + (y - 2 )^2 = 36

So. the center is (-1/2, 2) and the radius is 6

CPhill  Jul 10, 2014
Sort:

#1
+26550
+11

Alan  Jul 10, 2014
#2
+84238
+11

Alan has an interesting approach.......mine is a little different........but it should still work out the same...

Let's divide the original equation by 4....so we have

x^2 + x + y^2 - 4y - 127/4 =0          add  127/4 to both sides

x^2 + x + y^2 - 4y = 127/4                now, complete the sqaure on x and y

(x^2 + x + 1/4) + (y^2 - 4y + 4) =  127/4 + 1/4 + 4         simplifying, we have

(x + 1/2)^2  + (y - 2)^2  = 144/4

(x + 1/2)^2 + (y - 2 )^2 = 36

So. the center is (-1/2, 2) and the radius is 6

CPhill  Jul 10, 2014
#3
+26550
+8

Always good to have a couple of different approaches (even better when they give the same answers!).

Alan  Jul 10, 2014
#4
+84238
+3

Yep....unlike the "keycode" problem..... in which the number of different answers might actually be approaching the same limit as the number of actual keycodes......(LOL!!!)

CPhill  Jul 10, 2014

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