Consider the following circle. (If an answer does not exist, enter DNE.)
x2 + y2 − 10x + 4y + 15 = 0
(a) Find the x-intercepts.
(b) Find the y-intercept.

Question

Consider the following circle. (If an answer does not exist, enter DNE.)
x2 + y2 − 10x + 4y + 15 = 0
(a) Find the x-intercepts.
(b) Find the y-intercept.

anonymous · Answer

Solve the two quadratic equations, one with x=0, the other with y=0.

anonymous · Answer

To find the x-intercept, plug 0 into y and solve for x. 
To find the y-intercept, plug 0 into x and solve for y.

anonymous · Answer

Found those solutions yet. @tymiller ?

anonymous · Answer

i put them into the equations and my online hw ste said the answers were wrong

anonymous · Answer

trying it again

anonymous · Answer

What solutions did you come up with?

anonymous · Answer

(x^2-10x-25)+(y^2+4y-4)=radical 14
x=10+/- radical 10^2-4(1)(-25)/2            
x=5+/-5 radical 2
y=-4+/-radical 4^2-4(1)(-4)/2

zepdrix · Answer

Hmm

anonymous · Answer

y=-2+/-4 radical 2

zepdrix · Answer

(x-5)^2 + (y+2)^2 = 14

Is this the equation of the circle you were able to get it to?

anonymous · Answer

Yes, hmmm

zepdrix · Answer

Because I see that you were able to complete the square on x and y, after you do that, write it that way :D

zepdrix · Answer

the x term should have a +25, not -25 to complete that square :D minor detail

anonymous · Answer

From
$$x^2 + y^2 − 10x + 4y + 15 = 0$$
Y-intercepts are where:
$$y^2 + 4y + 15 = 0$$
X-intercepts are where:
$$x^2− 10x + 15 = 0$$
Neither are factorable, so use quadratic formula.

anonymous · Answer

Completing the square isn't necessary, but can be useful.
It would make it easier to graph, which can help visualize the solutions.

anonymous · Answer

still wont except the answer.

anonymous · Answer

From completing the square, you can see that the circle has center at (5,-2) with a radius of √14. The graph looks like this:
|dw:1348955659797:dw|

anonymous · Answer

yea i got that part

anonymous · Answer

From the graph, you see that you'll get two solutions for the x-intercepts, and no real solutions for the y-intercepts.
(again, completing the square and graphing is not necessary; use quadratic formula on the two equations for x and y.)

anonymous · Answer

:(  Webassign still doesnt like the answers

zepdrix · Answer

Hmm what did you put in? :)

zepdrix · Answer

$$x=\sqrt14+5$$$$x=-\sqrt14+5$$
Are you getting these for solutions? That's what I came up with :d lemme know if yer still confused.

anonymous · Answer

those answers do not work i have tried

zepdrix · Answer

Hmmmmm