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The circle enclosed in the square HIJK above touches only two points

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The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 09 Sep 2018, 07:55
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A
B
C
D
E

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  25% (medium)

Question Stats:

73% (01:20) correct 27% (01:00) wrong based on 44 sessions

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The circle enclosed in the square HIJK above touches only two points on the perimeter of the square, marked X and Y, and both are equidistant from point K. What is the radius of the circle?

(1) The sum of segment XK and segment YK is 16.
(2) The length of the radius is two-fifths the length of any side of the square.


Attachment:
image010.jpg
image010.jpg [ 4.49 KiB | Viewed 658 times ]

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Re: The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 09 Sep 2018, 09:12
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If the image is in scale answer is D


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Re: The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 09 Sep 2018, 09:32
The answer is E , bcos

Statement 1, => xk=8, yk=8 since x and y are equidistant from k, but it cannot be used to find the radius. - insufficient

Statement 2 => radius = (2/5)(side) but length of 1 side is not given - insufficient

Both Together, cannot give the length of the side or the radius.

Hence E
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The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 09 Sep 2018, 23:57
Bunuel wrote:
Image
The circle enclosed in the square HIJK above touches only two points on the perimeter of the square, marked X and Y, and both are equidistant from point K. What is the radius of the circle?

(1) The sum of segment XK and segment YK is 16.
(2) The length of the radius is two-fifths the length of any side of the square.


Attachment:
image010.jpg



From the given information , we can easily conclude that the given tangents KY and KX form a square if we just connect the point of contacts from circle's center .
So if we know the length of the KY or KX , we can find the radius . [KY=KX=Radius] .

From statement 1 , we can find the radius => Sufficient
Statement 2 => not sufficient

Answer is A
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The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 10 Sep 2018, 01:34
1
Bunuel wrote:
Image
The circle enclosed in the square HIJK above touches only two points on the perimeter of the square, marked X and Y, and both are equidistant from point K. What is the radius of the circle?

(1) The sum of segment XK and segment YK is 16.
(2) The length of the radius is two-fifths the length of any side of the square.


Attachment:
The attachment image010.jpg is no longer available


Question stem:- Radius=OX=OY=?

St1:- The sum of segment XK and segment YK is 16.
XK+YK=16
Or, XK=YK=8 (Since the circle touches the square at its adjacent sides , hence XK=YK)
Notice that OY=XK, so OY=8 (Please refer the attachment)
Similarly, OX=YK, so, OX=8
Sufficient.

St2:- The length of the radius is two-fifths the length of any side of the square.
Since measure of side of square is not known, hence radius can't be determined.
Insufficient.

Ans. (A)
Attachments

Figure square.jpg
Figure square.jpg [ 18.54 KiB | Viewed 447 times ]


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Re: The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 10 Sep 2018, 12:29
PKN wrote:
Bunuel wrote:
Image
The circle enclosed in the square HIJK above touches only two points on the perimeter of the square, marked X and Y, and both are equidistant from point K. What is the radius of the circle?

(1) The sum of segment XK and segment YK is 16.
(2) The length of the radius is two-fifths the length of any side of the square.


Attachment:
image010.jpg


Question stem:- Radius=OX=OY=?

St1:- The sum of segment XK and segment YK is 16.
XK+YK=16
Or, XK=YK=8 (Since the circle touches the square at its adjacent sides , hence XK=YK)
Notice that OY=XK, so OY=8 (Please refer the attachment)
Similarly, OX=YK, so, OX=8
Sufficient.

St2:- The length of the radius is two-fifths the length of any side of the square.
Since measure of side of square is not known, hence radius can't be determined.
Insufficient.

Ans. (A)



Nowhere does it say that X is at 90 degrees from O. For all we know, couldn't the center be slightly towards the left from X?
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Re: The circle enclosed in the square HIJK above touches only two points  [#permalink]

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New post 11 Sep 2018, 18:45
rajudantuluri wrote:
PKN wrote:
Bunuel wrote:
Image
The circle enclosed in the square HIJK above touches only two points on the perimeter of the square, marked X and Y, and both are equidistant from point K. What is the radius of the circle?

(1) The sum of segment XK and segment YK is 16.
(2) The length of the radius is two-fifths the length of any side of the square.


Attachment:
image010.jpg


Question stem:- Radius=OX=OY=?

St1:- The sum of segment XK and segment YK is 16.
XK+YK=16
Or, XK=YK=8 (Since the circle touches the square at its adjacent sides , hence XK=YK)
Notice that OY=XK, so OY=8 (Please refer the attachment)
Similarly, OX=YK, so, OX=8
Sufficient.

St2:- The length of the radius is two-fifths the length of any side of the square.
Since measure of side of square is not known, hence radius can't be determined.
Insufficient.

Ans. (A)



Nowhere does it say that X is at 90 degrees from O. For all we know, couldn't the center be slightly towards the left from X?


Hi rajudantuluri,

Given, sides of the square, HK and JK , touch the circle at the points X and Y respectively, implies that H-X-K and K-Y-J are the tangents.
DEFINITION OF Tangent:
A line passing a circle and touching it at just one point.The tangent line is always at the 90 degree angle (perpendicular) to the radius of a circle.
You know radius is the distance from the center to any point on the circle.

Therefore the perpendicular lines XO and YO meet at the center of circle.

Hope it helps.
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Regards,

PKN

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Re: The circle enclosed in the square HIJK above touches only two points &nbs [#permalink] 11 Sep 2018, 18:45
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