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How many points of intersection does the curve x^2 + y^2 = 4

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

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New post 13 Sep 2010, 13:26
1
19
00:00
A
B
C
D
E

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Question Stats:

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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\((2-y)^2 + y^2 = 4\)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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New post 13 Sep 2010, 13:45
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\((2-y)^2 + y^2 = 4\)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 x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)


Image

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 right-angled triangle whose other sides are of length x-a and y-b.

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, \(y=2-x\) is the equation of a line with x-intercept (2, 0) and y-intercept (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
graph.PNG [ 16.8 KiB | Viewed 7457 times ]


Answer: C (2).
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Re: did not understand the explanation  [#permalink]

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New post 13 Sep 2010, 14:00
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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New post 13 Sep 2010, 16:16
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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New post 13 Sep 2010, 16:50
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: \(y=x-2\), substitute \(x\) by \(y\) (or vise-versa) 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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New post 27 Sep 2010, 06:19
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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New post 11 Mar 2018, 20:40
Hi All,

If you want to do the work, then 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 the equations to answer this question, 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.

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Re: How many points of intersection does the curve x^2 + y^2 = 4 &nbs [#permalink] 11 Mar 2018, 20:40
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