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Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
 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
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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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
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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.
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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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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 ? THEORYIn 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^2For more check: math-coordinate-geometry-87652.htmlBACK TO THE ORIGINAL QUESTIONDoes 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. Answer: B.
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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
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 = 16y = 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) >= 0or Is (a^2 - 16) <= 0Is -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>16Doesn'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)
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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
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 = 16y = 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) >= 0or Is (a^2 - 16) <= 0Is -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>16Doesn'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
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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
13 Feb 2013, 21:27
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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.
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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ? [#permalink]
15 Feb 2013, 02:27
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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Re: Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?
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15 Feb 2013, 02:27
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