 Hi and welcome to the session. Let's work out the following question. The question says PAB is a sequence to a circle intersecting it at A and B and PT is a tangent to the circle. Prove that PA cross PB is equal to PT square. Use the above in the following. Two circles intersect each other at A and B. The common chord AB is produced to meet common tangent PQ to the circle at D. Prove that DP is equal to DQ. So let PAB be the sequence to this circle intersecting this circle at the point A and point B. PT is a tangent to the circle. So let us start with a solution to this question. First of all we do some construction here. What we do is we draw OM perpendicular to AB. We join OP OT OA. So like this we draw perpendicular to AB then we have joined O to P, O to T and O to A. Now we see the proof M is perpendicular to AB by construction. Therefore to be M because perpendicular drawn from the center of the circle at call bisects the call. So this will be equal to this and this we call 1. So let us write down the reason why does this happen. This happens because perpendicular drawn we center the circle bisects. The LHS of what we are supposed to prove LHS is P A into M minus AM. So P M minus AM multiplied by PB. PB is from 1 into we have P A into PB is equal to P M minus AM multiplied by P M plus AM because B M is same as AM. This is equal to M minus AM square. Now we see that P M is same as OP square minus OM square square root. So this is equal to OP square minus OM square minus AM square. This happens because this is the right angle triangle and from the Pythagoras theorem we have P M square will be equal to OP square minus OM square. This is further equal to OP square minus. Now OM square plus OA square is equal to sorry it is OM square plus AM square is equal to OA square. So we write here OA square. This is equal to OP square minus OT square because OA is equal to OT because they are the radii of the same circle. Now this is further equal to and this is equal to PT square. Because in right angle triangle OT P that is this triangle OP square is equal to PT square plus OT square. Therefore we can say that P A cross PB is equal to PT square. Now this is what we were supposed to prove in first part of the question. Now in second part of the question we are given two circles that intersect at A and B. Now AB is a common chord. PQ is a common tangent to both the circles. When AB is produced then it meets the tangent PQ at the point D. So let us see the proof now. Here we have to proof is equal to DQ. So let us see the proof. We see that for the bigger circle that is this one. PD is a tangent and DAB is a secant therefore DP square is equal to DA into DB. This we call one and this we get from also we see that for the smaller circle DQ is a tangent and DAB is a secant therefore DQ square is equal to DA into DB. This we call two and this also we get from above theorem equal to DQ square because both of them they are equal to DA into DB. Also we can say that DP is equal to DQ. So this is what we were supposed to prove in second part of the question. I hope that you understood the solution and enjoyed the session. Have a good day.