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Determine whether the graph of the given equation is a circle,a point or the null set.(choose any)
1.xsquare+ysquare+2x-2y+1=0
2.xsquare+ysquare+4x+8y-10=0

2006-08-15 00:30:29 · 18 answers · asked by gen 2 in Science & Mathematics Mathematics

18 answers

1) (x² + 2x + 1) + (y² -2y + 1) = -1 + 1 + 1
(x + 1)² + (y - 1)² = 1
is a circle with center (-1, 1) and radius 1.

2) (x² + 4x + 4) + (y² + 8y + 16) = 10 + 4 + 16
(x + 2)² + (y + 4)² = 30
is a circle with center (-2, -4) and radius √30.

2006-08-15 00:40:26 · answer #1 · answered by Anonymous · 1 0

1.
x² + y² + 2x - 2y + 1 = 0

2.
x² + y² + 4x + 8y - 10 = 0

The general circle (or point or null set) is
x² + y² + ax + by + c = 0
The center is (h,k) or (-a/2,-b/2)
The radius is sqrt(h² + k² - c) = 1/2 sqrt(a² + b² - 4c)
Now if the radius is positive then the equation is a circle.
If the radius is zero, then the equation is a point.
If the radius is negative, then the equation is a null set.

For the first equation
r = 1/2 sqrt(2² + (-2)² - 4(1)) = 1/2 sqrt(4 + 4 - 4) = 1/2 sqrt4 = 1/2 · 2 = 1
The first equation is a circle

For the 2nd equation
r = 1/2 sqrt(4² + 8² - 4(-10) = 1/2 sqrt(16 + 64 + 40) = 1/2 sqrt 120) = positive number
the 2nd equation is also a circle.

^_^
^_^

2006-08-15 00:43:12 · answer #2 · answered by kevin! 5 · 2 0

#1: x² + 2x + y² - 2y + 1 = 0 → x² + 2x + 1 + y² - 2y + 1 - 1 = 0 → (x+1)² + (y-1)² = 1

Thus this is a circle, centered at (-1, 1) with a radius of 1

#2: x² + 4x + y² + 8y - 10 = 0 → x² + 4x + 4 + y² + 8y + 16 - 30 = 0 → (x+2)² + (y+4)² = 30

This is a circle, centered at (-2, -4), with a radius of √30

2006-08-15 00:41:02 · answer #3 · answered by Pascal 7 · 1 0

xsquare+ysquare+2x-2y+1=0

is circle with center (-1,1) radius is one


xsquare+ysquare+4x+8y-10=0

is circle with center (-2,-4) radius is square root of 30

2006-08-15 00:47:19 · answer #4 · answered by !_! 2 · 0 0

xsquare+ysquare+4x+8y-10=0

2006-08-15 01:08:05 · answer #5 · answered by hard headed diabectic 2 · 0 0

I answered correctly. I got 10 points. I don't get a circle, a point or a null set.

2006-08-15 00:38:04 · answer #6 · answered by Anonymous · 0 0

the graph of the first equation is a circle:
(x+1)^2+(y-1)^2=1 mid piont (-1,1), r=1

the graph of the second equation is a circle:
(x+2)^2+(y+4)^2=30 mid point (-2,-4), r=sqrt(30)

2006-08-15 00:56:38 · answer #7 · answered by amir11elad 2 · 0 0

That's not true. Several people can answer correctly, however only one (even an incorrect answer) can get 10 points.

PS

write xsquare as "x^2"

2006-08-15 17:50:27 · answer #8 · answered by Anonymous · 1 0

10 points

2006-08-15 00:36:12 · answer #9 · answered by Anonymous · 1 0

equations of a circle with
1.centre (-1,1) and radius=1 unit
2.center (-2,-4) and radius(30)^1/2 units

2006-08-15 00:36:39 · answer #10 · answered by raj 7 · 1 0

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