Question

Find the center (h,k) and radius (r) of the circle:
2x^2+20x_2y^2=0

Answer 1

** 2x^2 +20x+2y^2=0

Answer 2

do you know what the general equation for a circle is?

Answer 3

no not really, I missed this chapter of class

Answer 4

\[\it is: (x-h)^{2} + (y-k)^{2} = r^{2}\]

the point (h, k) is the circle's center. and r is the radius

Answer 5

to put it in this form you'll need to complete the square for x.

completing the square: \[x^{2} + bx = (x+ (b/2))^{2} - (b/2)^{2}\]

Answer 6

so \[2x^{2} + 20x = 2(x^{2} + 10x) = 2[ (x + 5)^{2} - 25 ]\]

Answer 7

we now have: \[2(x + 5)^{2} + 2y^{2} = 50\]

Answer 8

\[= (x + 5)^{2} + y^{2} = 25\]

Answer 9

where did the 2 go

Answer 10

i divided both sides by two. it is equivalent


from the general equation, can you tell me what the center and radius are?

let me know when you want me to tell you

Answer 11

(h,k) = (5,1) r = 25?

Answer 12

(h, k) = (-5,0) r = sqrt25 = 5

Answer 13

why is it the square root and not 25^2?

Answer 14

the general equation has r^2 on the right side. meaning that r is really the sqrt of r^2

Answer 15

oh ok! So, to graph this I would put a point at (-5, 0) and go out 5 spaces ?

Answer 16

yes :)

Answer 17

yay!!

Answer 18

proof : http://www.wolframalpha.com/input/?i=2%28x%2B5%29^2+%2B+2y^2+%3D+50

glad i could help :)

Answer 19

so, if it touches (0,0) on the graph would that be an intercept or would (-10,0) be the only one?

Answer 20

the point (0,0) would be considered both an x and y intercept

Answer 21

Ok, so when I graph it and it asks for the intercepts I can say (-10, 0) and (0,0)

Answer 22

yes

Answer 23

yay :) thank you so much

Answer 24

glad i could help :)