If three circles having radii 1, 2, and 3 respectively lie o

Using statement 1, make an equilateral triangle with height \(2\sqrt{3}\) which is a little more than 3.4 since \(\sqrt{3}\) is a little more than 1.7. The side of this triangle will be 4 (since \(Height = (\sqrt{3}/2) * Side = 2*\sqrt{3}\))

On any one vertex, we will have the circle with the radius 2. On another, you will have to draw a circle with radius 3 but since the length of the side is only 4, this circle will intersect the circle with radius 2. Hence, statement 1 alone is sufficient.

Statement 2 doesn't tell us where the circles lie. They could lie very close to each other and intersect or very far from each other and not intersect. This statement alone is not sufficient.


Answer (A)

*Edited
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A.

height = 2sqrt3
hypoteneous = side of the eq. triangle = 4.

so clearly not.

Can you explain the answer? Not sure how the height and the side of a triangle relate in finding whether the circles intersect

A

Interesting question: does "osculation" means "intersection"? I guess yes. There is one point that belong to both circles. am I right?

A

stat 1:
since h = 2 sqrt(3), each side = 4
If the circles don't intersect, then one of the sides will be 3+2 = 5 which is not possible. Therefore, we can determine that at least 2 of the circles intersect. Suff.

Stat 2:
This doesn't tell us anything. They could or could not intersect. Insuff.

Bunuel,

Is their better way to approach this question?

From 1)
Let "a" be the side of the Equilateral triangle.
Given height = 2*3= 6 . ( height of equilater triangle with side a is sqrt(3) * a/2)
==> a= 4* sqrt(3).
therefore the distance between the centers of all the three circles ,taken two at a time is = 4 * sqrt(3) .
which clearly shows that the circles don't itersect.
( 4*sqrt(3) is greater than the all possible sum of the radii of the circles taken two at a time 2+3 , 1+2 , 1+3 = 5,3,4. ) Sufficient


From 2) The distance between two circles is less than 6 , which implies it may be greater than 5 or less than 5.
We are not sure the circles will intersect or not . Insufficient
Hence A

If three circles having radii 1, 2, and 3 respectively lie on a plane, do any two of these circles intersect?
(1) Centers of the circles form an equilateral triangle with height 2√3.
(2) Centers of the circles do not lie on the same line.

Hi, can anyone show how geometry looks like for St(1), please.

OE

Merging topics. Please refer to the discussion above.
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Hi Karishma,

The length of the side of the triangle would be 4, and not 2sqrt(3). This is the height as specified in the question. However the answer does not vary.

Santora

Hi VeritasPrepKarishma,

As per you "length of the side is only 3.4" but according to me the lenght of side should be 4.

formulae-

Altitude of equilateral triangle= (\sqrt{3}/2) * (side)

2\sqrt{3} = (\sqrt{3}/2) * (side)

So, side= 4..

Where I am wrong..??
Please assist.
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Yes, you are right. The height is given so the side will be 4. You will not be able to draw the circles with radii 2 and 3 without intersecting. The answer stays the same.
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I have been posting my queries but i am not receiving any reply to the same. Can someone please help me with my doubts. From the statement 1 by finding out the length of the side for equilateral triangle how can we say that they are not intersecting??

You get that the side of the equilateral triangle is 4. The centres of the 3 circles lie on the three vertices of the triangle.
On any one vertex, we will have the circle with the radius 2. On another, you will have to draw a circle with radius 3 but since the length of the side is only 4, this circle will intersect the circle with radius 2. Hence two of these circles will intersect.
Hence, statement 1 alone is sufficient.
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