Question

GEOMETRY HELP!!


The general form of the equation of a circle is Ax2 + By2 + Cx + Dy + E = 0, where A = B 0. If the circle has a radius of 3 units and the center lies on the y-axis, which set of values of A, B, C, D, and E might correspond to the circle?

Answer 1

I always thought that the formula for the circle is \[(x-h)^2+(y-k)^2=r^2\]

Answer 2

yeah that's what i thought but i think the islander is referring to the gen eqtn of a conic

Answer 3

you can forget answer 1, since it has \(A=0\) so what i guess you have to do, which is rather annoying, is find the center and radius of the other 4 circles and see which one fits the conditions

Answer 4

first one would be
\[x^2+y^2+8x+9=0\] complete the square via
\[x^2+8x+y^2=-9\\
(x+4)^2+y^2=-9+16=7\] and you can stop right there because the radius of that one is \(\sqrt7\) not \(3\)

Answer 5

It's the second one, D=0 because it is on the y-axis.

Answer 6

no i don't think so

Answer 7

i just found the radius of the second one to be \(\sqrt7\)

Answer 8

the radius is sqrt of E silly. Nothing else.

Answer 9

really? hmmm

Answer 10

C and D are just expressing the offset of the circle from the origin.

Answer 11

wonder how i got \(\sqrt7\) then ...

Answer 12

lets try the third one
\[x^2+y^2-8y=7=0\] complete the square via
\[x^2+y^2-8y=-7\\
x^2+(y-4)^2=-7+19=9\] center is \((0,4)\) and radius is \(\sqrt9=3\) (notice it is not \(\sqrt7\) !!)

Answer 13

typo there, should have been
\[x^2+(y-4)^2=-7+16=9\] but conclusion is correct

Answer 14

so it is B, huh

Answer 15

to me, it's C

Answer 16

yes it's C

Answer 17

because the co-efficient C=0

Answer 18

it is always C

Answer 19

the formula for a circle whose center (0,k) is x^2 + (y-k)^2 = 9 ( the given information)
so, x^2 + y^2 -2ky +k^2 -9 =0 compare to Ax^2 + By^2+ Cx+Dy + E =0
make comparison, A =1 , B = 1, C =0 , we can stop here because there is only one option satisfy those ,it's C

Answer 20

alright, thanks everybody for the help. I really do appreciate it