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

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

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10 Feb 2005, 13:07
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Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16
(2) a=|b|+5
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Apr 2012, 13:20, edited 1 time in total.
Edited the question and added the OA

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11 Apr 2012, 13:22
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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: $$x^2+y^2=r^2$$

For more check: math-coordinate-geometry-87652.html

BACK TO THE ORIGINAL QUESTION

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.

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

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12 Apr 2012, 11:14
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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.

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17820 [14], given: 235 Director Joined: 19 Nov 2004 Posts: 554 Kudos [?]: 280 [7], given: 0 Location: SF Bay Area, USA ### Show Tags 10 Feb 2005, 16:18 7 This post received KUDOS 2 This post was BOOKMARKED Another way of doing this: (x-a)^2 + (y-b)^2=16 intersect the y-axis when x=0 => a^2 + y^2 + b^2 - 2yb =16 or y^2 -2yb + a^2 + b^2 -16 =0 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. Kudos [?]: 280 [7], given: 0 VP Joined: 18 Nov 2004 Posts: 1430 Kudos [?]: 49 [2], given: 0 ### Show Tags 10 Feb 2005, 14:32 2 This post received KUDOS I think it's "B". 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 Kudos [?]: 49 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7738 Kudos [?]: 17820 [1], given: 235 Location: Pune, India Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink] ### Show Tags 13 Feb 2013, 21:27 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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27 Oct 2017, 05:40
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ayas7 wrote:
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.

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:
$$(0-a)^2 + (y-b)^2 = 16$$
$$a^2 + (y - b)^2 = 16$$
$$(y - b)^2 = 16 - a^2$$
$$(y - b) = \sqrt{16 - a^2}$$
$$y = b + \sqrt{16 - a^2}$$
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11 Apr 2012, 13:14
nocilis wrote:
Another way of doing this:
(x-a)^2 + (y-b)^2=16 intersect the y-axis when x=0

=> a^2 + y^2 + b^2 - 2yb =16
or
y^2 -2yb + a^2 + b^2 -16 =0

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 ?

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

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13 Feb 2013, 21:15
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.

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

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15 Feb 2013, 02:27
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From the given problem, we will have the curve intersect the y-axis iff we get a real value for the given curve at x=0.

Thus, putting x=0, we get

$$(-a)^2+(y-b)^2 = 16$$

$$or (y-b)^2 = 16-a^2$$

Now for real value of y, we have to have $$16-a^2>=0.$$

$$or a^2<=16$$

$$-4<=a<=4.$$

F.S 1 not sufficient.

F.S 2, as a= mod(b)+5, the minimum value of a has to be 5. Thus sufficient. The curve will not intersect the y-axis.

B.
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12 Jun 2013, 04:34
banerjeea_98 wrote:
I think it's "B".

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.

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12 Jun 2013, 05:37
dyuthi92 wrote:
banerjeea_98 wrote:
I think it's "B".

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 $$a = |b| + 5=6>4$$.

Check here: does-the-curve-x-a-2-y-b-2-16-intersect-the-y-axis-14046.html#p1072781 or here: does-the-curve-x-a-2-y-b-2-16-intersect-the-y-axis-14046.html#p1073168

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

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18 Jun 2013, 10:27
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

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.

but took more than 2 minutes to solve

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

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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, 08:47
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.

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

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24 Oct 2017, 11:19
nocilis wrote:
Another way of doing this:
(x-a)^2 + (y-b)^2=16 intersect the y-axis when x=0

=> a^2 + y^2 + b^2 - 2yb =16
or
y^2 -2yb + a^2 + b^2 -16 =0

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 "

did you use b^2-4ac?

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

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24 Oct 2017, 11:27
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.

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!

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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?   [#permalink] 24 Oct 2017, 11:27
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