Tangents are drawn to the circle x^2+y^2=50 from a point P lying on the x-axis.These tangents meet the y-axis at points P1 and P2.Possible coordinates of P so that area of triangle PP1P2 is minimum,is/are (A)(10,0) (B)(10root2,0) (C)(-10,0) (D)(-10root2,0) Correct option is A,C

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We have:S:x2+y2-50=0Let co-ordinates of P be x1,y1=p,0Equation of pair of tangents from  x1,y1 is:SS1=T2x2+y2-50x12+y12-50=xx1+yy1-502x2+y2-50p2+0-50=px-502x2+y2-50p2-50=px-502This pair of tangents meet y-axis at poitns where x=0y2-50p2-50=0-502y2-50=2500p2-50y2=2500p2-50+50=2500+50p2-2500p2-50y2=50p2p2-50y=±50p2p2-50=±50p2-50pHence We have Q0,50p2-50p and R0,-50p2-50pNote that:Q and Rare equidistant from origin.Using distance formula you will get QR asQR=2p50p2-50 which will be base of triangle.Height of triangle will be OP=pArea=A=12×Base×HeightA=12×2p50p2-50×pA=p250p2-50A2=50p4p2-50Differentiate with respect to p, we get:2A.dAdp=200p3p2-50-50p42pp2-502For maxima/minima, dAdp=0200p3p2-50-50p42pp2-502=0200p3p2-50-50p42p=050p34p2-50-2p2=04p2-50-2p2=04p2-200-2p2=02p2=200p2=100p=±10I leave checking double derivative for you.Hence coordinates ofP are 10,0 or -10,0OptionA,C

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