Question

Find the x and y intercept of the circle x^2-10x+y^2+2x=-17

Answer 1

I meant x^2-10x+y^2+2y=-17

Answer 2

you know what a "perfect square trinomial" is, right?

Answer 3

(x-5)^2+(y+1)^2=9
I got that far, but I ran into imaginary numbers when substituting x for 0 to find the y-intercept.

Answer 4

well, let's just start by grouping only
and then we'll "complete the square" by using our very good friend, Mr Zero

\(\bf x^2-10x+y^2+2y=-17 \implies (x^2-10x)+(y^2+2y)=-17 \)

what would be the 3rd number in each grouped binomial, to get us a "perfect square trinomial"?

Answer 5

25 and 1, respectively.
25+1=26
Add the same number to both sides, which results in agjnpggg=9

Answer 6

quantity, rather.

Answer 7

yeap :)

Answer 8

you add and subtract, we're just borrowing from 0, so +somevalue -somevalue = 0

Answer 9

ohhh... scratch that

Answer 10

\(\bf x^2-10x+y^2+2y=-17 \implies (x^2-10x)+(y^2+2y)=-17\\
(x^2-10x+5^2)+(y^2+2y+1^2)=-17 \\\implies (x-5)^2+(y+1)^2 - 5^2-1^2 = 17\\\quad \\
(x-5)^2+(y+1)^2 = 9 \implies (x-5)^2+(y+1)^2 = 3^2\)

Answer 11

Yep.

Answer 12

hmm, anyhow, I missed a -17 there, ....\(\bf x^2-10x+y^2+2y=-17 \implies (x^2-10x)+(y^2+2y)=-17\\
(x^2-10x+5^2)+(y^2+2y+1^2)=-17 \implies (x-5)^2+(y+1)^2 - 5^2-1^2 = -17\\\quad \\
(x-5)^2+(y+1)^2 = 9 \implies (x-5)^2+(y+1)^2 = 3^2
\)

Answer 13

so that's the circle's center and radius

Answer 14

(5,-1)=Center
3=Radius.
Yup.

Answer 15

yeap

Answer 16

But x and y intercepts ;-;

Answer 17

well, to get the "x" intercept, set y =0, solve for "x"

to get the "y" intercept, set x =0, solve for "y"

Answer 18

... I guess I didn't read careful enough... though you needed the equation, anyhow, it's easier to get the intercepts with it

Answer 19

(x−5)^2+(y+1)^2=9

(0−5)^2+(y+1)^2=9
25+(y+1)^2=9
(y+1)^2=-16

Now what?

Answer 20

hmm, well... that will give an imaginary value

Answer 21

I think you would reject it.
(x−5)^2+(0+1)^2=9

(x−5)^2=8
x=sqrt(8)+5

But it's really ugly, so I didn't like it.

Answer 22

notice the center of the circle, (5, -1)


meaning, the circle never touches the y-axis, thus the imaginary value for the y-intercept

Answer 23

-->But it's really ugly, so I didn't like it. <---- hehhe
well, it doesn't have to be good-looking, it just have to be valid =)

Answer 24

Right, then. Thanks.