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I'm not sure the reasoning behind the red part in bold section......how did you get to 4b^2 - 4(1)(a^2 + b^2 -16) >=0 ?

THEORY In an x-y Cartesian 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

Does the curve (x - a)^2 + (y - b)^2 = 16 intersect the Y axis?

Curve of (x - a)^2 + (y - b)^2 = 16 is a circle centered at the point (a, \ b) and has a radius of \sqrt{16}=4. Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis. So basically the question asks whether |a|>4: if it is, then the answer will be NO: the curve does not intersect with Y axis and if it's not, then the answer will be YES: the curve intersects with Y axis.

(1) a^2 + b^2 > 16 --> clearly insufficient as |a| may or may not be more than 4.

(2) a = |b| + 5 --> as the least value of absolute value (in our case |b|) is zero then the least value of a will be 5, so in any case |a|>4, which means that the circle does not intersect the Y axis. Sufficient.

Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
12 Apr 2012, 10:14

8

This post received KUDOS

Expert's post

christoph wrote:

Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16 (2) a=|b|+5

What do we mean by 'intersect the y axis'? We mean that the x co-ordinate is 0.

So let's put x = 0 and see what we get: (0-a)^2 + (y-b)^2 = 16 y = b + (16 - a^2)^{1/2}

Now what decides whether we get a value for y or not? Obviously, if (16 - a^2) is negative, y will have no real value and the curve will not intersect the y axis. If instead (16 - a^2) is 0 or positive, the curve will intersect the y axis. So we can answer the question asked if we know whether (16 - a^2) is positive/0 or not.

Is (16 - a^2) >= 0 or Is (a^2 - 16) <= 0 Is -4 <= a <= 4?

We have simplified the question stem as much as we could. Let's go on to the given statements.

(1) a^2+b^2>16 Doesn't tell us about the value of a. a could be 2 or 10 or many other values.

(2) a=|b|+5 a is 5 or greater since |b| is at least 0. This tells us that 'a' does not lie between -4 and 4. Hence this statement is sufficient to answer the question.

For this quedratic eqn. to have real roots,
4b^2 - 4(1)(a^2 + b^2 -16) >=0
or a^2 <=16

(1) a^2+b^2>16

It does not say anything about a^2 being <=16. So can't say

(2) a=|b|+5

|b| is postive or zero. So a is atleast 5 and a^2 is atleast 25.
This contradicts our requirement that a^2 <=16. So no, it does not intersect the y axis

Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
13 Feb 2013, 20:27

1

This post received KUDOS

Expert's post

Sachin9 wrote:

Hi Karishma, What does intersect mean? In our Indian education system , we have learnt that intersection means that the arcs/lines cross each other. I remember reading in MGMAT forum that on the GMAT intersection would also mean that the line may be a tangent to the arc. . So in that case, even if |a|=4, then the circle intersects the y axis .. Is my understanding correct?

Regards, Sachin

Yes, intersect (as far as GMAT is concerned) means touch. x co-ordinate is 0 there. Where do the lines/curves go after touching is immaterial. Notice that in the explanation above, I have given: "If instead (16 - a^2) is 0 or positive, the curve will intersect the y axis."

It means that even if a = 4, the curve will intersect the y axis. _________________

For the circle to intersect, we have to have a center which is 4 units or < away from the y axis as 4 is the radius of this circle.

I. a2+b2 > 16.....a = 3, b = 8, circle intersect, if a = 5 and b = 0, it doesn't....insuff

II. a = |b|+5.....means that x coordinate of center is always +ve and >=5.....any center with >=5 as X coordinate will never intersect Y axis as the circle radius is 4....suff

For this quedratic eqn. to have real roots, 4b^2 - 4(1)(a^2 + b^2 -16) >=0or a^2 <=16

(1) a^2+b^2>16

It does not say anything about a^2 being <=16. So can't say

(2) a=|b|+5

|b| is postive or zero. So a is atleast 5 and a^2 is atleast 25. This contradicts our requirement that a^2 <=16. So no, it does not intersect the y axis

B it is.

I'm not sure the reasoning behind the red part in bold section......how did you get to 4b^2 - 4(1)(a^2 + b^2 -16) >=0 ?

Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
13 Feb 2013, 20:15

VeritasPrepKarishma wrote:

christoph wrote:

Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16 (2) a=|b|+5

What do we mean by 'intersect the y axis'? We mean that the x co-ordinate is 0.

So let's put x = 0 and see what we get: (0-a)^2 + (y-b)^2 = 16 y = b + (16 - a^2)^{1/2}

Now what decides whether we get a value for y or not? Obviously, if (16 - a^2) is negative, y will have no real value and the curve will not intersect the y axis. If instead (16 - a^2) is 0 or positive, the curve will intersect the y axis. So we can answer the question asked if we know whether (16 - a^2) is positive/0 or not.

Is (16 - a^2) >= 0 or Is (a^2 - 16) <= 0 Is -4 <= a <= 4?

We have simplified the question stem as much as we could. Let's go on to the given statements.

(1) a^2+b^2>16 Doesn't tell us about the value of a. a could be 2 or 10 or many other values.

(2) a=|b|+5 a is 5 or greater since |b| is at least 0. This tells us that 'a' does not lie between -4 and 4. Hence this statement is sufficient to answer the question.

Answer (B)

Hi Karishma, What does intersect mean? In our Indian education system , we have learnt that intersection means that the arcs/lines cross each other. I remember reading in MGMAT forum that on the GMAT intersection would also mean that the line may be a tangent to the arc. . So in that case, even if |a|=4, then the circle intersects the y axis .. Is my understanding correct?

Regards, Sachin _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

For the circle to intersect, we have to have a center which is 4 units or < away from the y axis as 4 is the radius of this circle.

I. a2+b2 > 16.....a = 3, b = 8, circle intersect, if a = 5 and b = 0, it doesn't....insuff

II. a = |b|+5.....means that x coordinate of center is always +ve and >=5.....any center with >=5 as X coordinate will never intersect Y axis as the circle radius is 4....suff

what if b is less than zero. as we know |x| = x if x <0 and |x| = -x if x >0. in this case we get two equations for a 1) a-b = 5( a could be 9 and be could be 4. second equation for a is a+b = 5 (then b=0 and a=5 or a=5 and b=0 or a=1 and b =4..)could be possibilities if a=5.it say it does intersect y axis but in all the other three cases it intersects with y axis.in this way i got E. please tell me if am missing something.

For the circle to intersect, we have to have a center which is 4 units or < away from the y axis as 4 is the radius of this circle.

I. a2+b2 > 16.....a = 3, b = 8, circle intersect, if a = 5 and b = 0, it doesn't....insuff

II. a = |b|+5.....means that x coordinate of center is always +ve and >=5.....any center with >=5 as X coordinate will never intersect Y axis as the circle radius is 4....suff

what if b is less than zero. as we know |x| = x if x <0 and |x| = -x if x >0. in this case we get two equations for a 1) a-b = 5( a could be 9 and be could be 4. second equation for a is a+b = 5 (then b=0 and a=5 or a=5 and b=0 or a=1 and b =4..)could be possibilities if a=5.it say it does intersect y axis but in all the other three cases it intersects with y axis.in this way i got E. please tell me if am missing something.

If b<0, for example, if b=-1, then still a = |b| + 5=6>4.

Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
19 Jun 2013, 07:24

For what it's worth, it's not as if you have to solve EVERY problem in under two minutes. That's just the average. This is a particularly difficult 700+ question!

s1: put a= 0 and b =5 then the cirlce touches y axis but if the center is 3, 8 then the cirlce never touches y axis so, not sufficient

s2: this clearly shows, a is not equat to 0 so, it never touches y axis so sufficient.

This question asks us if the circle intersects the y-axis, so how is the x-axis relevant?

Thanks!

Bunuel wrote:

catty2004 wrote:

I'm not sure the reasoning behind the red part in bold section......how did you get to 4b^2 - 4(1)(a^2 + b^2 -16) >=0 ?

THEORY In an x-y Cartesian 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

Does the curve (x - a)^2 + (y - b)^2 = 16 intersect the Y axis?

Curve of (x - a)^2 + (y - b)^2 = 16 is a circle centered at the point (a, \ b) and has a radius of \sqrt{16}=4. Now, if a, the x-coordinate of the center, is more than 4 or less than -4 then the radius of the circle, which is 4, won't be enough for curve to intersect with Y axis. So basically the question asks whether |a|>4: if it is, then the answer will be NO: the curve does not intersect with Y axis and if it's not, then the answer will be YES: the curve intersects with Y axis.

(1) a^2 + b^2 > 16 --> clearly insufficient as |a| may or may not be more than 4.

(2) a = |b| + 5 --> as the least value of absolute value (in our case |b|) is zero then the least value of a will be 5, so in any case |a|>4, which means that the circle does not intersect the Y axis. Sufficient.

Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
06 Jul 2014, 22:00

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