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# DS: Curve (m09q03)

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DS: Curve (m09q03) [#permalink]  07 Jan 2009, 15:45
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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. $$a = |b| + 5$$

I have no idea how to tackle this... please explain. Thanks.

[Reveal] Spoiler: OA
B

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Re: DS: Curve (m09q03) [#permalink]  22 Dec 2010, 02:39
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321kumarsushant wrote:
E is final answer.

both statement cant answer the question.

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.

Answer: B.
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Re: DS: Curve (m09q03) [#permalink]  15 Dec 2009, 04:09
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Hi,

You guys have to tell me what exactly is not clear in my explanation above.

You have to keep in mind that we are dealing with an equation of a circle with radius 4. $$a$$ places the center of this circle closer to or farther from the Y axis. From S2 we know that $$a \ge 5$$, so the circle does not intersect the Y axis.

OA is B.
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Re: DS: Curve [#permalink]  07 Jan 2009, 20:36
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bigfernhead 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

I have no idea how to tackle this... please explain. Thanks.

Agree with E.

Equation of the circle: (x-a)^2 + (y-b)^2 = 16
Origin = (a, b) & redius of this circle = 4

The values of (a, b) doesnot decide the position of circle. they are x and y along with a and b determin the circle's location.

We need to know
1: a^2 + b^2 > 16
a and b could be anything because their squares are +ve and > 16.

2: a = lbl + 5.
now we know that: a is +ve and is > 5 but b? we do not know. it (b) could be anything.

1&2: are we getting anything extra bit of infromation from 2 on 1 or vice versa. No.

so E.
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Re: DS: Curve (m09q03) [#permalink]  22 Dec 2010, 04:15
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Bunuel wrote:
321kumarsushant wrote:
E is final answer.

both statement cant answer the question.

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.

Answer: B.

YOUR explanation is very detailed and appreciable.

consider stmnt 2.
take a case as
a=2 & b=3
or b=3 & a=2. both cond is in agreement with the statement 2.
so the eq will be (x-2/3)^2+(x-3/3)^2=16
this circle doesn't intersects the Y axis anywhere.

so the final ans will be E.
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Re: DS: Curve (m09q03) [#permalink]  22 Dec 2010, 04:58
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321kumarsushant wrote:
YOUR explanation is very detailed and appreciable.

consider stmnt 2.
take a case as
a=2 & b=3
or b=3 & a=2. both cond is in agreement with the statement 2.
so the eq will be (x-2/3)^2+(x-3/3)^2=16
this circle doesn't intersects the Y axis anywhere.

so the final ans will be E.

First of all, OA (the final answer) for this question is B, not E.

Next, your examples (the red part) is not valid as if $$a=2$$ and $$b=3$$ then $$2\neq{|3|+5=8}$$ or if $$a=3$$ and $$b=2$$ then $$3\neq{|2|+5=7}$$ so these values of $$a$$ and $$b$$ do not satisfy statement (2).

Moreover, if $$a=2$$ and $$b=3$$ then circle with equation $$(x - a)^2 + (y - b)^2 = 16$$ does intersect Y-axis (so your conclusion is also wrong, because the formula you wrote is wrong).

Again the circle given by $$(x-a)^2+(y-b)^2=r^2$$ has center at point (a, b) and radius r, so if $$a=2$$ and $$b=3$$ then circle given by $$(x - 2)^2 + (y - 3)^2 = 16$$ has its center at (2,3) and has a radius of 4 and it intersects Y axis at points $$(0, \ 3-2\sqrt{3})$$ and $$(0, \ 3+2\sqrt{3})$$. You can see it on the below diagram:
Attachment:

MSP30219de6acg5dc779ic0000351eaf739i21c8f1.gif [ 3.04 KiB | Viewed 9420 times ]

Hope it helps.
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Re: DS: Curve (m09q03) [#permalink]  22 Dec 2010, 01:47
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E is final answer.

both statement cant answer the question.
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Re: DS: Curve (m09q03) [#permalink]  10 Jun 2011, 00:45
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bblast wrote:
Bunuel, I donno if u r still on these forums, but I need help with this :

" so in any case $|a|>4$, which means that the circle does not intersect the Y axis."

I understood everything else except this statement.

$$|a|>4$$ i.e either a>4 OR a<-4.
If a, the x coordinate of the center of the circle, is more than 4 units away from the y-axis, the circumference of the circle will NEVER intersect the y-axis because the radius of the circle is 4 units. Thus, if we know that a is indeed >4 or <-4, we will definitely know that the circle doesn't intersect y-axis.
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Re: DS: Curve [#permalink]  07 Jan 2009, 18:01
(x-a)^2+(y-b)^2 = 16
is a circlce with centre(a,b) and radius 4.

(1)=>distance of (a,b) from origin in the coordinate system is>4. There are multiple circles with their center on the x-axis which will not intersect y-axis. Then again there are circles with center on the y-axis which will intersect y-axis. hence, not sufficient.
(2)=>a=modb+5
Thus, the circle has center at (5+b,b) or (5-b,b)
I solved the equn of circle with the y-axis (x=0).
we get
y-b = sqrt(b^2-10b-9) or sqrt(b^2+10b-9)
in any case it is possible to have imaginary solution for some values of b and not others.
hence, not sufficient.
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Re: DS: Curve [#permalink]  07 Jan 2009, 20:17
I am in doubt.
from 2) a= |b|+5

I take that to mean a is always positive and >= 5. With a radius of 4 and x cooridnate >=5 it cannot intersect the y axis.

2 should be suff.

Please correct me if my reasoning is wrong.
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Re: DS: Curve [#permalink]  08 Jan 2009, 17:24
you are right!
The center of the circle is atleast 5 units away from the origin; so no intersection with x=0 is possible. I wonder why I could not derive it from my equation solving though.
GMAT TIGER wrote:
bigfernhead 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

I have no idea how to tackle this... please explain. Thanks.

Agree with E.

Equation of the circle: (x-a)^2 + (y-b)^2 = 16
Origin = (a, b) & redius of this circle = 4

The values of (a, b) doesnot decide the position of circle. they are x and y along with a and b determin the circle's location.

We need to know
1: a^2 + b^2 > 16
a and b could be anything because their squares are +ve and > 16.

2: a = lbl + 5.
now we know that: a is +ve and is > 5 but b? we do not know. it (b) could be anything.

1&2: are we getting anything extra bit of infromation from 2 on 1 or vice versa. No.

so E.

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Re: DS: Curve (m09q03) [#permalink]  25 Aug 2009, 20:32
how is possible that if radius is 4, then a^2 + b^2 >16. What this eq is telling me is contradicting with the question. Isn't it?
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Re: DS: Curve (m09q03) [#permalink]  26 Aug 2009, 05:35
Intersect with Y axis means x=0, y has a value.

so we can get a^2+(y-b)^2=16
(y-b)^2=16-a^2>=0

then just need
|a|<=4

neither 1), 2) nor combining them, can't get |a|<=4

Ans: E
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Re: DS: Curve [#permalink]  27 Sep 2009, 11:38
GMAT TIGER wrote:
Equation of the circle: (x-a)^2 + (y-b)^2 = 16

Question states curve. Is this a standard equation with which we can infer that the curve is nothing but a circle?

Also, OA is given as B.

Can someone explain this?
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Re: DS: Curve [#permalink]  27 Sep 2009, 13:24
I totally agree with the explanation below. The Y axis is at X=0. Since from S2 $$a \ge 5$$, we can be sure that X coordinate of the center of the circle will be 5 points away from the Y axis. If the radius of the circle is 4 points, we are sure that the circle does NOT intersect the Y axis. We should not be concerned with what the value of $$b$$ is as we have to check only the intersection with the Y axis. We would have to check the value of $$b$$ only if the question asked for the intersection with the X axis where Y=0.

The "curve" here refers to the equation of the circle.

I hope the explanation makes sense. The OA is B.
nfernandes wrote:
I am in doubt.
from 2) a= |b|+5

I take that to mean a is always positive and >= 5. With a radius of 4 and x coordinate >=5 it cannot intersect the y axis.

2 should be suff.

Please correct me if my reasoning is wrong.

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Re: DS: Curve (m09q03) [#permalink]  11 Dec 2009, 11:30
I go with choice E. But please someone explain let me know the correct answer and the necessary explanation.
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Re: DS: Curve (m09q03) [#permalink]  11 Dec 2009, 12:18
I don't understand the answer to this. Wouldn't the circle not intercept the y-axis if a is larger than 16?
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Re: DS: Curve (m09q03) [#permalink]  22 Dec 2010, 09:47
thanks buddy...
i was a bit confused, but it got over .
thank you.
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Re: DS: Curve (m09q03) [#permalink]  09 Jun 2011, 21:58
Bunuel, I donno if u r still on these forums, but I need help with this :

" so in any case $|a|>4$, which means that the circle does not intersect the Y axis."

I understood everything else except this statement.
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Re: DS: Curve (m09q03) [#permalink]  10 Jun 2011, 10:31
+1 fluke

my brains' light finally flickered after ur comment. got it now.
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Re: DS: Curve (m09q03)   [#permalink] 10 Jun 2011, 10:31

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# DS: Curve (m09q03)

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