Q2 - mock test 1 pending


Question 2:

Let ABCD be a square of side length 5. A circle passing through A is tangent to segment CD at T and meets AB and AD again at X≠AX \ne A and Y≠AY \ne A respectively given that XY=6XY = 6. Find AT2AT^2.

Solution:

Let O be the center of the circle, and let Z be the foot from O to AD.
Since XY is a diameter, OT = ZD = 3, so AZ = 2. Then OZ=5OZ = \sqrt{5} and
AT=OZ2+22=30AT = \sqrt{OZ^2 + 2^2} = \sqrt{30}.



Another solution:

OT is perpendicular to DC. Extend OT so that it intersects AB at P.
OAX is isosceles since OA = OX = radius = 3.
AP = PX = sqrt(5).
Triangle ATP is right angled at P.
AT^2 = AP^2 + TP^2 = 30.
TP = 5 = side length of the square.

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