Curve Question

Question

This one got me. Any help would be appreciated.

How many points of intersection does curve x^2 + y^2 = 4 have with x + y = 2?

0
1
2
3
4

prude_sb · Answer

we can rule out D & E since a line can instersect a circle only in 2 pts....

The points of intersection will satisfy both equations x^2 + y^2 = 4 and x + y = 2


so x = y - 2

=> (y-2)^2 + y^2 = 0
2y^2 - 4y + 4 = 0
y^2 - 2y + 2 = 0
solve quadratic eq to get values of x,y
if discriminant is >0 then we have 2pts of intersection 
if discriminant is =0 then we have 1pt of intersection (line is a tangent to the circle)
if discriminant is <0 then we have 0pts of intersection (line does not intersect the circle)

in this case discriminant = 4 - 8 = -4 so ans is 0 (A)

crrrrg · Answer

here's another way, i realized (after a while) that x^2 +y^2 =4 is the equation for a circle with radius=2 and centered at the origin.

https://www.mathwarehouse.com/geometry/c ... circle.php

x+y=2 is the equation of a line and it won't touch the circle more than twice so you can eliminate the last two. once you note that one is a circle and the other a line i think the next easiest step is to draw a fast coordinate plane in the notepad and see if the line passes through or is tangent to the circle, since the equation for this particular line and circle are simple. The other way is to solve both for y, knowing that the maximum number of intersections is two.

y1=2-x

y2=SQRT(4-x^2)

At x=0, y1=y2=2.
At x=2, y1=y2=0.

So the line passes through the circle twice, at (0,2) and at (2,0).

sgoll · Answer

I think it should be x=2-y

Also , The equation of the given circle is x^2+Y^=4

What do you say?

Fig · Answer

Well, it's not to me

... But I'm passing here

Notice that : (2-y)^2 = ((-1)*(y-2))^2 = (-1)^2*(y-2)^2 = (y-2)^2

For the other part.... U are right... 4 is missing

sgoll · Answer

I agree (y-2)^=(2-y)^2
Further, the eqn when equals 4, it results to answer to C: 2
Explanation
substituting x=2-y in the eqn of the Circle we get
(2-y)^2+y^2=4
Solving for y we get y(y-2)=0
=> either y=0 or y=2 giving two points of intersection (0,2) and (2,0)

Fig · Answer

Yes

... Absolultly

bmwhype2 · Answer

OA is C.

Two roots yields two coordinates ===> 2 points of intersection.

lawsohn · Answer

How'bout my solution? 
In x + y = 2, we see y-intercept =(0,2), x-intercept =(2,0)
In the curve that is centered by the origin, we know it passes (0,2), (-2,0), (0,-2), and (2,0). 
So, the line and the curve meet at two points.

liftoff · Answer

Got two points sub equ 2 into 1 and determine with b^2 - 4ac

Spidy001 · Answer

x^2 + y^2 = 4 is a circle with radius 2.
x+y = 2 is a straight line , with slope -1 , x intercept = 2 , y intercept = 2.
=> point of intersection = (2,0) and (0,2) = 2 points

Answer is C.

pike · Answer

No need to get too bogged down on the maths here. We have a circle and a line, so we are either going to have zero intersections, one intersection or two intersections.

Sketch out the circle, sketch the line. The line intersects x and y at 2, as does the circle. Two intersections.

gaurav2k101 · Answer

me too first got A then C

samramandy · Answer

It was a simple one.

A circle and a line intersection. Either 0, 1 or 2 points of intersection. Rest has been explained very well by others.

BR
Mandeep

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