Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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) >=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

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 ?

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.

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.

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

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________