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 ≠ A X \ne A and Y ≠ A Y \ne A respectively given that X Y = 6 XY = 6 . Find A T 2 AT^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 O Z = 5 OZ = \sqrt{5} and A T = O Z 2 + 2 2 = 30 AT = \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.