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M16-04

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Math Expert
Joined: 02 Sep 2009
Posts: 52905

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15 Sep 2014, 23:58
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Difficulty:

35% (medium)

Question Stats:

72% (00:40) correct 28% (00:48) wrong based on 78 sessions

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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

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Joined: 02 Sep 2009
Posts: 52905

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15 Sep 2014, 23:58
Official Solution:

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

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 $$(0, 2)$$.

Alternative Explanation

Look at the diagram below:

$$x^2 + y^2 = 4$$ is the equation of a circle centered at the origin and with the radius of 2. Now, $$y=2-x$$ is the equation of a line with $$x$$-intercept $$(2, 0)$$ and $$y$$-intercept $$(0,2)$$. Notice that these points are also the intercepts of given circle with $$X$$ and $$Y$$ axis hence at these points the line and the circle intersect each other.

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Joined: 08 Feb 2014
Posts: 204
Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41
WE: Analyst (Commercial Banking)

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03 Nov 2014, 17:55
Is there a quick way to recognize the equation of a circle?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 52905

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04 Nov 2014, 04:25
2
JackSparr0w wrote:
Is there a quick way to recognize the equation of a circle?

Thanks

Yes.

In an xy coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
$$(x-a)^2+(y-b)^2=r^2$$

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: $$x^2+y^2=r^2$$

For more check: math-coordinate-geometry-87652.html
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Concentration: Finance
GMAT 1: 650 Q39 V41
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04 Nov 2014, 15:36
Manager
Joined: 10 Sep 2014
Posts: 82
GPA: 3.5
WE: Project Management (Manufacturing)

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25 Mar 2018, 22:34
1
Bunuel wrote:
Official Solution:

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

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 $$(0, 2)$$.

Alternative Explanation

Look at the diagram below:

$$x^2 + y^2 = 4$$ is the equation of a circle centered at the origin and with the radius of 2. Now, $$y=2-x$$ is the equation of a line with $$x$$-intercept $$(2, 0)$$ and $$y$$-intercept $$(0,2)$$. Notice that these points are also the intercepts of given circle with $$X$$ and $$Y$$ axis hence at these points the line and the circle intersect each other.

Bunuel I did not understand this part "$$(2 - y)^2 + y^2 = 4$$ 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 $$(0, 2)$$." like where did this come from "[m](2 - y)^2 + y^2 = 4" ?
Math Expert
Joined: 02 Sep 2009
Posts: 52905

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25 Mar 2018, 22:56
1
Bunuel wrote:
Official Solution:

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

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 $$(0, 2)$$.

Alternative Explanation

Look at the diagram below:

$$x^2 + y^2 = 4$$ is the equation of a circle centered at the origin and with the radius of 2. Now, $$y=2-x$$ is the equation of a line with $$x$$-intercept $$(2, 0)$$ and $$y$$-intercept $$(0,2)$$. Notice that these points are also the intercepts of given circle with $$X$$ and $$Y$$ axis hence at these points the line and the circle intersect each other.

Bunuel I did not understand this part "$$(2 - y)^2 + y^2 = 4$$ 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 $$(0, 2)$$." like where did this come from "(2 - y)^2 + y^2 = 4" ?

First of all, check this post: https://gmatclub.com/forum/m16-184073.html#p1437884

We are given that $$x^2 + y^2 = 4$$ have with line $$x + y = 2$$. From $$x + y = 2$$ we have $$x = 2 - y$$. Substitute $$x = y - 2$$ into $$x^2 + y^2 = 4$$ to get $$(2 - y)^2 + y^2 = 4$$.
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Joined: 10 Sep 2014
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26 Mar 2018, 05:59
Another query Bunuel how x^2+y^2=4 curve is a circle with radius 4? Thanks.
Re: M16-04   [#permalink] 26 Mar 2018, 05:59
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