SOLUTION: Find the intercepts of the circle (x+1)2 + (y-4)2=9

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Question 650185: Find the intercepts of the circle (x+1)2 + (y-4)2=9
Found 3 solutions by lynnlo, ewatrrr, MathLover1:
Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
x and y intercepts
15/2~7.5

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Find the intercepts of the circle 

(x+1)2 + (y-4)2=9 = 3^2   |Radius of 3

NO x-intercepts, when x = 0, y = 4 ± sqrt(8), (0,1.17) and (0, 6.83)

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
%28x+-+h%29%5E+2+%2B+%28y+-+k%29%5E+2+=+r+%5E2
you have %28x%2B1%29%5E2+%2B+%28y-4%29%5E2=9
so, h=-1, k=4, and r%5E2=9...->r=3
center at C%28h+%2C+k%29+=+C%28-1+%2C+4%29+ and radius r+=+3
To find the x-intercepts, set y+=+0 in the above equation and solve
for x.
%28x%2B1%29%5E2+%2B+%280-4%29%5E2=9
%28x%2B1%29%5E2+%2B+%28-4%29%5E2=9
%28x%2B1%29%5E2+%2B16=9
%28x%2B1%29%5E2+=9-16
%28x%2B1%29%5E2+=-7
sqrt%28%28x%2B1%29%5E2+%29=sqrt%28-7%29
x%2B1+=sqrt%28-7%29
x+=sqrt%28-7%29-1
as you can see, there is none of x-+intercepts+ because there is no real solution
To find the y-+intercepts, set x+=+0 in the above equation and solve
for y.
%280%2B1%29%5E2+%2B+%28y-4%29%5E2=9
%281%29%5E2+%2B+%28y-4%29%5E2=9
1+%2B+%28y-4%29%5E2=9
%28y-4%29%5E2=9-1
%28y-4%29%5E2=8
sqrt%28%28y-4%29%5E2%29=sqrt%288%29
y-4=sqrt%288%29.........sqrt%288%29=2.8 and sqrt%288%29=-2.8
y=2.8%2B4
y=6.8
and
y=-2.8%2B4
y=1.2

y-+intercepts+ are 6.8 and 1.2

drawing%28300%2C300%2C-5%2C10%2C-5%2C10%2Cgrid%281%29%2Ccircle%28-1%2C4%2C3%29%29