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Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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Updated on: 11 Apr 2012, 13:20
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Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ? (1) a^2+b^2>16 (2) a=b+5
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Originally posted by christoph on 10 Feb 2005, 13:07.
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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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 xy Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that: \((xa)^2+(yb)^2=r^2\) 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 rightangled triangle whose other sides are of length xa and yb. If the circle is centered at the origin (0, 0), then the equation simplifies to: \(x^2+y^2=r^2\) For more check: mathcoordinategeometry87652.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 xcoordinate 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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Another way of doing this:
(xa)^2 + (yb)^2=16 intersect the yaxis 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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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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nocilis wrote: Another way of doing this: (xa)^2 + (yb)^2=16 intersect the yaxis 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 (xa)^2 + (yb)^2=16 intersect the yaxis ?
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12 Apr 2012, 11:14
christoph wrote: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
(1) a^2+b^2>16 (2) a=b+5 What do we mean by 'intersect the y axis'? We mean that the x coordinate is 0. So let's put x = 0 and see what we get: \((0a)^2 + (yb)^2 = 16\) \(y = b + (16  a^2)^{1/2}\) 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)
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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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13 Feb 2013, 21:15
VeritasPrepKarishma wrote: christoph wrote: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
(1) a^2+b^2>16 (2) a=b+5 What do we mean by 'intersect the y axis'? We mean that the x coordinate is 0. So let's put x = 0 and see what we get: \((0a)^2 + (yb)^2 = 16\) \(y = b + (16  a^2)^{1/2}\) 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
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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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13 Feb 2013, 21:27
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 coordinate 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 (xa)^2 + (yb)^2=16 intersect the yaxis ?
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15 Feb 2013, 02:27
From the given problem, we will have the curve intersect the yaxis iff we get a real value for the given curve at x=0. Thus, putting x=0, we get \((a)^2+(yb)^2 = 16\) \(or (yb)^2 = 16a^2\) Now for real value of y, we have to have \(16a^2>=0.\) \(or a^2<=16\) \(4<=a<=4.\) 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 yaxis. B.
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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) ab = 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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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) ab = 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: doesthecurvexa2yb216intersecttheyaxis14046.html#p1072781 or here: doesthecurvexa2yb216intersecttheyaxis14046.html#p1073168Hope it helps.
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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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18 Jun 2013, 10:27
christoph wrote: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
(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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Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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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 xcoordinate 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: (xa)^2 + (yb)^2 = 16 Let a = b = 5 and we are trying to find y intercept => x = 0 25 + (y5)^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 yaxis. This sort of makes sense from the diagram as x coordinate of the center determines the distance of center of circle from yaxis and y coordinate decides the distance of center of circle from xaxis.
Summary: In (xa)^2 + (yb)^2 = c^2 if: (i) a > c => This distance of center of circle from yaxis is MORE THAN the radius. So no point of intersection between circle and Yaxis. (ii) b > c => This distance of center of circle from xaxis is MORE THAN the radius. So no point of intersection between circle and xaxis.



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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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24 Oct 2017, 11:19
nocilis wrote: Another way of doing this: (xa)^2 + (yb)^2=16 intersect the yaxis 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^24ac?



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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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24 Oct 2017, 11:27
VeritasPrepKarishma wrote: christoph wrote: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
(1) a^2+b^2>16 (2) a=b+5 What do we mean by 'intersect the y axis'? We mean that the x coordinate is 0. So let's put x = 0 and see what we get: \((0a)^2 + (yb)^2 = 16\) \(y = b + (16  a^2)^{1/2}\) 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!



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Re: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
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27 Oct 2017, 05:40
ayas7 wrote: VeritasPrepKarishma wrote: christoph wrote: Does the curve (xa)^2 + (yb)^2=16 intersect the yaxis ?
(1) a^2+b^2>16 (2) a=b+5 What do we mean by 'intersect the y axis'? We mean that the x coordinate is 0. So let's put x = 0 and see what we get: \((0a)^2 + (yb)^2 = 16\) \(y = b + (16  a^2)^{1/2}\) 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: \((0a)^2 + (yb)^2 = 16\) \(a^2 + (y  b)^2 = 16\) \((y  b)^2 = 16  a^2\) \((y  b) = \sqrt{16  a^2}\) \(y = b + \sqrt{16  a^2}\)
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