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by crackgmat007 » Wed Nov 04, 2009 5:00 pm
How many points of intersection does the curve x^2 + y^2 = 4 have with line x + y = 4?

oa - 2

Can someone solve this using equations
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by gmatv09 » Wed Nov 04, 2009 10:51 pm
x^2+y^2 = 4 is the eqn of the circle with center @ origin (0,0) and radius = 4

y = -x + 4 is a line with y-intercept of 4 and a -ve slope of -1.
The line intercepts the x-axis at (4,0)

Hence the answer is 2
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by NikolayZ » Wed Nov 04, 2009 10:57 pm
x^2+y^2=4,
x+y=4.

I think you know the pattern of the first equation.
It is cercle with the center at (0;0), and radius 4.

The second one is the line, y=4-x, if y=0, then x=4, and if x=0, then y=4. Then we have 2 intersect points, hence the answer would be indeed 2.
Well, i tried to solve this one algebraically, but i got discriminant that is less then 0... Literally it means those eqs. don't have common roots :) Then i thought it would be less time consuming to solve this one graphically....)

GMATV09, hehe)))
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by ayushr » Thu Nov 05, 2009 1:42 am
gmatv09 and NikolayZ -
Isn't x^2+y^2 = 4 a circle with radius 2 ( Not 4)and center at (0,0) ??
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by NikolayZ » Thu Nov 05, 2009 1:50 am
ooops, you are right Ayushr. My school math is 7 years away so i made this silly mistake. indeed the radius must be 2.
Then the answer to this particular problem is 0.
If the circle equation looks like : x^2+y^2=16(radius 4), then there are 2 common roots.

P.s. Then it is clear why the discriminant is negative and these line and circle have no common roots.
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by crackgmat007 » Thu Nov 05, 2009 12:07 pm
Guys OA is 2. We are missing something. This is a GMAT clubs problem.
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