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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.

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]

Show Tags

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.

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]

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07 Aug 2016, 19:47

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

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23 Oct 2016, 07:47

Hello Friends, I would like to clarify 1 point which wasn't clear to me: Bunuel "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." Here's why: Equation: (x-a)^2 + (y-b)^2 = 16 Let a = b = 5 and we are trying to find y intercept => x = 0 25 + (y-5)^2 = 16 y^2 - 10y + 25 + 9 = 0 y^2 - 10y + 34 = 0 b^2 - 4ac = 100 - 4*(1)*(34) = 100 - 136 = -36 delta < 0 => NO solution. Thus curve [ y^2 - 10y + 34 = 0 ] doesn't intersect y-axis. This sort of makes sense from the diagram as x co-ordinate of the center determines the distance of center of circle from y-axis and y co-ordinate decides the distance of center of circle from x-axis.

Summary: In (x-a)^2 + (y-b)^2 = c^2 if: (i) |a| > c => This distance of center of circle from y-axis is MORE THAN the radius. So no point of intersection between circle and Y-axis. (ii) |b| > c => This distance of center of circle from x-axis is MORE THAN the radius. So no point of intersection between circle and x-axis.

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

B it is.

Can you explain how you calculated this please? "For this quedratic eqn. to have real roots, 4b^2 - 4(1)(a^2 + b^2 -16) >=0 or a^2 <=16 "

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

Show Tags

24 Oct 2017, 10:27

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)

can you please share the steps of simplification with: (0−a)2+(y−b)2=16(0−a)2+(y−b)2=16 y=b+(16−a2)1/2?

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)

can you please share the steps of simplification with: (0−a)2+(y−b)2=16(0−a)2+(y−b)2=16 y=b+(16−a2)1/2?

thank you!

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