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11. a circle centred at C on the x-axis touches a straight line L1:y =x.Another line L2 perpendular toL1 touches the circle at E. OB=1a. show that triangleACO = triangle ACDb. find OC and the radiusc.find the equation of L2d find ABe. find the coordinates of B and Ehttp://homelf.kimo.com.tw/keraope/kk.JPG 顯示更多 11. a circle centred at C on the x-axis touches a straight line L1:y =x.Another line L2 perpendular toL1 touches the circle at E. OB=1 a. show that triangleACO = triangle ACD b. find OC and the radius c.find the equation of L2 d find AB e. find the coordinates of B and E http://homelf.kimo.com.tw/keraope/kk.JPG

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Sorry to say that your diagram is not good. Point O (origin) should lie on the line L1 because L1 : y=x. (a) Since, L1: y=x. angle AOC = 45degree angle ADC = 180 - angle A - angle AOC = 180 - 90 - 45 = 45 degree (given angle A = 90) So, Triangle OAD has two equal base angles. So, side OA = AD. AC bisects angle A (since AB and AE tangents to circle)) AC is common edge. So, triangle ACO is equivalent to triangle ACD (b) angle OBC = 90 because line OB is tangent to circle. angle BOC = 45 because slope of L1 = 1 angle BCO = 180 - 90 - 45 = 45 degree So, triangle OBC has two equal base angles. So, side OB = BC So, BC = 1 (given OB = 1) Radius = BC = 1 (c) All the interior angle of ABCE are right angles. (AB, AE is tangent to circle C and Angle A = 90) Since, ABCE is a rectangle. Opposite sides are equal. Now, BC = CE (radius of circle) So, ABCE is also a square! So, AB = BC = CE = AE So, AB = 1--------------(*). so, OA = OB + AB = 1 + 1 = 2. So, A = (2cos45, 2 sin45) = (sq. root 2, sq. root 2) Slope of L2 = -1 / slope of L1 = -1/1 = -1 L2 passes throught A. The equation of L2 is : y - sq root 2 = -1 ( x - sq root 2) x + y - 2* sq root 2 = 0 (d) AB = 1 (from above (*) ) (e) B = (1* cos 45, 1*sin45) = (1 / sq root 2, 1 / sq root 2) To show: OBEC is a parallelogram. Angle OBC = 45 Angle EDC = 180 - angle A - angle OBC = 180 - 90 -45 = 45 Angle ECD = 180 - angle CED - angle EDC = 180 - 90 - 45 = 45 So, OB is parallel to EC. OB = BC (based angles equal) EC = ED (based angles equal) OB = BC = EC Triangle OBC is equivalent to triangle ECD X-ordinate of point E = X-ordinate of point B + radius of circle Y-ordinate of point E = Y-ordinate of point B So, E = ( 1+ (1/sq root 2), 1/sq root 2). 2006-10-17 18:58:19 補充： (a)So, triangle ACO is equivalent to triangle ACD -- should be conguent not equivalent(e)Triangle OBC is equivalent to triangle ECD -- should be conguent not equivalent