SOLUTION: I am in intermediate algebra, my tutor is gone for the weekend and I need some assistance.
Problem: Find the center, radius, x-intercepts and y-intercepts of the circle X^2 + y^
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Question 376780: I am in intermediate algebra, my tutor is gone for the weekend and I need some assistance.
Problem: Find the center, radius, x-intercepts and y-intercepts of the circle X^2 + y^2 - 6x + 4y - 12 = 0
I am confused on the x-intercepts of my circle
This is what I have done so far:
X^2 - 6X + y^2 +4y = 12
x^2 - 6x + 9 + y^2 + 4y + 4 = 12 + 9 + 4
(x-3)^2 + (y + 2)^2 = 25
Y -intercepts:
(x-3)^2 + (y+2)^2 = 25
(0-3)^2 + (y+2)^2 = 25
9 + (y+2)^2 = 5^2
-9 + 9 + (y+2)^2 = 5^2 - 9
(y+2)^2 = 4
square it
y+2 = + or - 4
y = -6
y = 2
x intercepts
x^2 - 6x - 12 = 0
Now I am stuck - I can't figure out how to figure out the x intercept
I believe that vertex is : (3, -2)
radius is: (5)
y intercepts (0.-6) (0,2)
x intercepts (-2,0) except that my circle is not exactly on -2
Thank you for your assistance
Carol
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
^2\ =\ 16)
This is a circle, so it doesn't have a vertex. The center is
)
And the radius is 5.
Find the 
intercepts the same way you found the 
intercepts. Substitute 0 for and solve.
^2\ +\ (0\ +\ 2)^2\ =\ 25)
^2\ =\ 21)
Roughly 7.6 and -1.6
The intercept points that you have are correct.
The intercept points are
)
and
)
John
My calculator said it, I believe it, that settles it
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