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How many points of intersection does the curve x^2 + y^2 = 4 [#permalink]
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13 Sep 2010, 14:26
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How many points of intersection does the curve x^2 + y^2 = 4 have with line x+y =2 ? A. 0 B. 1 C. 2 D. 3 E. 4 did not understand the explanation. Curve \(x^2 + y^2 = 4\) is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives\((2y)^2 + y^2 = 4\)or \(2y^2  4y + 4 = 4\) or \(y^2  2y = 0\) from where y=0 and y=2. Thus, the line and the circle intersect at points (2,0) and (0,2).
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Re: did not understand the explanation [#permalink]
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13 Sep 2010, 14:45
seekmba wrote: did not understand the explanation.
How many points of intersection does the curve \(x^2 + y^2 = 4\) have with line x+y =2 ?
0 1 2 3 4
Curve \(x^2 + y^2 = 4\) is a circle with radius 2 and the center at the origin. A line cannot have more than 2 points of intersection with a circle, so we can eliminate choices D and E. To answer the question, we can either draw a sketch or solve the system of equations. The second approach gives\((2y)^2 + y^2 = 4\)or \(2y^2  4y + 4 = 4\) or \(y^2  2y = 0\) from where y=0 and y=2. Thus, the line and the circle intersect at points (2,0) and (0,2). Can you please specify what part of the solution didn't you understand? Anyway: In an xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\) This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a rightangled triangle whose other sides are of length xa and yb. If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\) So \(x^2 + y^2 = 4\) is the equation of circle centered at the origin and with radius equal to 2. Now, \(y=2x\) is the equation of a line with xintercept (2, 0) and yintercept (0,2) these points are also the intercepts of given circle with X and Y axis hence at these points line and circle intersect each other. Attachment:
graph.PNG [ 16.8 KiB  Viewed 3074 times ]
Answer: C (2).
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Re: did not understand the explanation [#permalink]
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13 Sep 2010, 15:00
Thanks so much Bunuel. I was not able to picture the solution given. It makes total sense now.



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Re: did not understand the explanation [#permalink]
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13 Sep 2010, 17:16
In circle and line, I beleieve answer could be just 1(tangent) or 2(line). I am not able to recollect what was the method to know if line is tanget to circle or not. Anyone?
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Re: did not understand the explanation [#permalink]
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13 Sep 2010, 17:50
saxenashobhit wrote: In circle and line, I beleieve answer could be just 1(tangent) or 2(line). I am not able to recollect what was the method to know if line is tanget to circle or not. Anyone? Actually there is a third case: when circle and line don't have any points of intersection. As for the solutions: well if it's an easy case, for example if line is \(y=2\) we can say that it's tangent to the circle right away or if line is \(y=5\) we can say right away that they don' intersect at all. For harder cases you can use the approach used in initial post: \(y=x2\), substitute \(x\) by \(y\) (or viseversa) in \(x^2+y^2=4\) and then solve for \(y\) (this value will be \(y\) coordinate of the intersection point(s)). If you'll get one solution for \(y\) it would mean that line is tangent to circle (as you'll get one point (x,y)), if you'll get two solutions for \(y\) it would mean that line has two intersection points with circle (as you'll get two points (x,y)) and if you'll get no solution for \(y\) it would mean that line has no intersection point with circle. Hope it's clear.
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Re: did not understand the explanation [#permalink]
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27 Sep 2010, 07:19
I think this should be included in the circles chapter of the math book



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Re: How many points of intersection does the curve x^2 + y^2 = 4 [#permalink]
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