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Re: Curve Question [#permalink]
22 Feb 2007, 21:14

bz9 wrote:

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

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)

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).

Re: Curve Question [#permalink]
23 Feb 2007, 07:00

Fig wrote:

sgoll wrote:

prude_sb wrote:

The points of intersection will satisfy both equations x^2 + y^2 = 4 and x + y = 2

so x = y - 2

I think it should be x=2-y

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

prude_sb wrote:

(y-2)^2 + y^2 = 0

What do you say?

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

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)

Re: Curve Question [#permalink]
23 Feb 2007, 07:11

sgoll wrote:

Fig wrote:

sgoll wrote:

prude_sb wrote:

The points of intersection will satisfy both equations x^2 + y^2 = 4 and x + y = 2

so x = y - 2

I think it should be x=2-y

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

prude_sb wrote:

(y-2)^2 + y^2 = 0

What do you say?

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

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)

How many points of intersection does curve x^2 + y^2 = 4 have with x + y = 2?

0 1 2 3 4

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.

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

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.

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