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M16-04

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M16-04  [#permalink]

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New post 16 Sep 2014, 00:58
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (00:40) correct 27% (00:47) wrong based on 77 sessions

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Re M16-04  [#permalink]

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New post 16 Sep 2014, 00:58
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:

Image

\(x^2 + y^2 = 4\) 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.


Answer: C
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Re: M16-04  [#permalink]

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New post 03 Nov 2014, 18:55
Is there a quick way to recognize the equation of a circle?

Thanks
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Re: M16-04  [#permalink]

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New post 04 Nov 2014, 05:25
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:
\((x-a)^2+(y-b)^2=r^2\)

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: \(x^2+y^2=r^2\)

For more check: math-coordinate-geometry-87652.html
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Re: M16-04  [#permalink]

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New post 04 Nov 2014, 16:36
Very helpful, thanks Bunuel!
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M16-04  [#permalink]

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New post 25 Mar 2018, 23:34
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:

Image

\(x^2 + y^2 = 4\) 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.


Answer: C
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" ?
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M16-04  [#permalink]

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New post 25 Mar 2018, 23:56
1
sadikabid27 wrote:
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:

Image

\(x^2 + y^2 = 4\) 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.


Answer: C
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 \((2 - y)^2 + y^2 = 4\).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M16-04  [#permalink]

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New post 26 Mar 2018, 06:59
Another query Bunuel how x^2+y^2=4 curve is a circle with radius 4? Thanks.
Re: M16-04 &nbs [#permalink] 26 Mar 2018, 06:59
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