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Bunuel

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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Bunuel

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 2939 times ]

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PKN
Bunuel

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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rajudantuluri
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Bunuel

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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