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by Brent@GMATPrepNow » Thu Jun 26, 2014 8:44 pm
How many points of intersection does the curve x² + y² = 4 have with line x + y = 2?
A) 0
B) 1
C) 2
D) 3
E) 4
The points of intersection will be points (x,y) that satisfy BOTH equations.

So, how many different solutions does this system have?
x² + y² = 4
x + y = 2

If x + y = 2 then x = 2 - y
Now take x² + y² = 4 and replace x with (2 - y) to get: (2 - y)² + y² = 4
Expand: 4 - 4y + y² + y² = 4
Simplify: 2y² - 4y = 0
Factor: 2y(y - 2) = 0
Solve: y = 0 or y = 2

If y = 0 (and x + y = 2), then x = 2. So, one point of intersection is (2, 0)
If y = 2 (and x + y = 2), then x = 0. So, another point of intersection is (0, 2)

So, we have 2 points of intersection in total.
Answer: C

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by [email protected] » Thu Jun 26, 2014 9:21 pm
Hi shibsriz,

If you want to do the work, the two equations in this question can be physically graphed, so that you can "see" the points of intersection.

You don't need to graph it, but you have to recognize the rarer graphing rule in this question:

The equation X^2 + Y^2 = 4 is a circle, centered around the Origin, with a radius of 2. The points (2,0), (0,2), (-2,0) and (0,-2) are all on the circumference of the circle

The equation X + Y = 2 can be rewritten as Y = -X + 2 and will be a straight line.

By definition, the line will intersect with the circle at either 0, 1 or 2 points.

Knowing the above 4 "obvious" points on the circle will allow you to quickly find 2 points on the line: (2,0) and (0,2). Then you can stop; a line can't intercept a circle at more than 2 points.

Final Answer: C

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by GMATinsight » Fri Jun 27, 2014 11:51 pm
If you plot the graph X^2 + Y^2 = 4, it's c circle with radius 2

If you plot the graph X + Y = 2, it's c Line with X-Intercept and Y-Intercept 2

The point of intersections are two as mentioned in the graph attached

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by GMATGuruNY » Sat Jun 28, 2014 5:42 am
How many points of intersection does the curve x² + y² = 4 have with line x + y = 2?
A) 0
B) 1
C) 2
D) 3
E) 4
One more approach:

x + y = 2
(x+y)² = 2²
x² + y² + 2xy = 4.

Substituting x² + y² = 4 into x² + y² + 2xy = 4, we get:
4 + 2xy = 4
2xy = 0
xy = 0.

Implication:
Either x=0 or y=0.
If x=0, then -- since x+y = 2 -- y=2.
If y=0, then -- since x+y = 2 -- x=2.
Thus, there are two points of intersection:
x=0, y=2
x=2, y=0.

The correct answer is C.
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