P(x, y) is the intersection point between the circle (x2 + y2 = 4) and

Question

P(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2). Which of the following can be the point P?

A. (1, 2).
B. (2, 0).
C. (0, -2).
D. (-2, 0).
E. (2, 2).

FacelessMan · Answer

A quick glance at the options, and we can see that A and E do not satisfy the circle equation. Now we are left with B, C and D. Now B and C do not satisfy the line equation. So we are just left with D. And thats the answer.

gauravsaggis1 · Answer

circle x^2+y^2=4 cuts line y=x+2.
so intercept will be x^2+(x+2)^2=4
x(x+1)=0, 
x=0 or x=-1.
 in the options only c has x=0. 
Ans C.

CounterSniper · Answer

x^2+x^2+4x+4=4
2x^2+4x=0
2x(x+2)=0
x=0
x=-2

alternatively

Among the given answer choices B,C,D will satisfy the equation of the circle
B and C however , will not satisfy equation of the line
correct answer is D

law258 · Answer

Simple plug and chug:

x^2 +(x+2)^2 = 4
x^2 + x^2 +4x+4 = 4
2(x^2) +4x=0
2x(x+2)=0
x =0 or -2

if x = 0 --> (plug into second equation) --> y = 2 --> (0,2); not an option, thus eliminate

if x=-2 --> y =0 --> (-2,0) --> D is the correct answer.

grichagupta · Answer

Alternate solution:
x^2+y^2 = 4 is a circle with centre at origin (0,0) and radius 2. Hence, (2,0) and (0,2) are two such points that lie on this circle.

Also, y=x+2 is a line that intersects the circle in IInd Quadrant and touches it at point (0,2) and (-2,0). Hence, quickly glancing through options such that x coordinate is negative and y positive, we get (-2,0) as one of the options. Hence, D

jahidhassan · Answer

Need more explanation for better understanding.

Psk13 · Answer

equation of line y=x+2 right?
also it is given circle and line intersect at point p
so at point p the any value of x,y which satisfies for the line ,will also satisfy the circle equation
only d option does that 
so correct answer is D

m i correct?  Bunuel

EMPOWERgmatRichC · Answer

Hi All,

This question can be solved by TESTing THE ANSWERS. Notice how each answer is a co-ordinate. We just have to find the one that 'fits' both equations and then we can stop working.

Answer A: (1,2) does NOT fit the first equation. Eliminate Answer A.
Answer B: (2,0) fits the first equation but NOT the second. Eliminate Answer B.
Answer C: (0,-2) fits the first equation but NOT the second. Eliminate Answer C.
Answer D: (-2,0) fits the first equation AND the second. This MUST be the answer.

Final Answer:  D

GMAT assassins aren't born, they're made,
Rich

coolkl · Answer

We are given two equations and two unknowns. So we start by substituting y = x+2 in the equation of circle
x^2 + (x+2)^2 = 4
x^2+x^2+4x + 4 = 4
x(2x+4) = 0
x = 0 or x = -2
If x = 0 then y = 2 and point is (0,2) using y = x+2
or 
If x = -2 then y = 0 and point is (-2,0)
From the options we see that Option D holds true.
Answer = D

vityakim@gmail.com · Answer

P(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2). Which of the following can be the point P?

A. (1, 2).
B. (2, 0).
C. (0, -2).
D. (-2, 0).
E. (2, 2).

My 2 cents.
I would just plug it in.
As P intersects two equations, the value of P must satisfy both.
Only D works.

JeffTargetTestPrep · Answer

Since P(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2), the coordinates of point P must fulfill both the equation of the circle and the line. Let’s test each answer choice.
 
A) (1, 2)
 
y = x + 2
 
2 = 1 + 2 ?
 
2 = 3 ?
 
Since 2 does not equal 3, answer choice A is not correct.
 
B) (2, 0)
 
y = x + 2
 
0 = 2 + 2 ?
 
0 = 4 ?
 
Since 0 does not equal 4, answer choice B is not correct.
 
C) (0, -2)
 
y = x + 2
 
-2 = 0 + 2 ?
 
-2 = 2 ?
 
Since -2 does not equal 2, answer choice C is not correct.
 
D) (-2, 0)
 
y = x + 2
 
0 = -2 + 2 ?
 
0 = 0 ? → Yes
 
x^2 + y^2 = 4
 
(-2)^2 + 0^2 = 4 ?
 
4 = 4 ? → Yes
 
Since the ordered pair (-2,0) fulfills both equations, answer choice D is correct.
 
Alternate Solution:
 
We have two equations and two unknowns, so we can substitute (x + 2) for y in the circle equation:
 
x^2 + (x + 2)^2 = 4
 
x^2 + x^2 + 4x + 4 = 4
 
2x^2 + 4x = 0
 
x^2 + 2x = 0
 
x(x + 2) = 0
 
x = 0 or x = -2
 
For x = 0, let’s substitute x = 0 into the linear equation y = x + 2, which yields y = 2; the ordered pair (0,2) is not an answer choice.
 
For x = -2, let’s substitute x = -2 into the linear equation y = x + 2, which yields y = 0; the ordered pair (-2,0) is answer choice D.
 
Answer: D

TheKingInTheNorth · Answer

P(x, y) is the intersection point between the circle (x^2 + y^2 = 4) and the line (y = x +2). Which of the following can be the point P?

A. (1, 2).
B. (2, 0).
C. (0, -2).
D. (-2, 0).
E. (2, 2).

I made a mistake . 
But this is how GMAT plays . Making a simple question complicated by putting an equation of circle . 
Note here we are given a circle and line equation .
Ask to find the point of intersection between circle and point .

Note that point will always satisfy equation of line and but it also satisfy the equation of circle .

You use both to eliminate the answer choices .

Hence D is the ans .

bumpbot · Answer

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P(x, y) is the intersection point between the circle (x2 + y2 = 4) and

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