 Hi and how are you all today? The question says if PAB is a secant to the circle, intersecting the circle at A and B and PT is the tangent segment proves that PAB into PB is equal to PT square. Using the above theorem, prove the following. Two circles intersect each other at PNQ. From point R on PQ produced two tangents RB and RC are drawn to the circle, to the two circles touching them at B and C. Then we need to prove that RB is equal to RC. So let us first start with the solution by proving the theorem which is given to us. My hair we are given PAB is the secant to a circle, center O, tangent to the circle. A into PB is equal to PT square. We need to join O to the midpoint that is of AB also B and T. We have joined O to M, A, B and T. This is a line from the center to the midpoint of a chord that means it is perpendicular to the chord. So let us start with our proof. O is the center of the circle, midpoint of the chord AB. So therefore we can write that OM is perpendicular on AB because center to the midpoint of the chord, perpendicular to the also we have PA is equal to minus AM but we know that AM is equal to MB because M is the midpoint. Therefore PB is equal to PM plus this be the first equation. Let this be the second equation. Now on multiplying first equation and the second equation we get. Left hand sides will get multiplied by left hand side. So we have PA into PB equal to PM minus AM into PM plus AM. So here A minus B equals B can be written as A square minus B square. Now we know that OM is perpendicular to the chord. So therefore in this triangle we can write that PM square is equal to OP square minus OM square that is by Pythagoras. AM square is equal to OA square minus OM square that is again by Pythagoras theorem thus we can write PA into PB is equal to in place of PM square now we can write OP square minus OM square and now in place of AM square we can write OA square minus OM square. On solving it has OP square minus OM square minus OA square plus OM square these two will get cancelled. So OP square minus OA square which can be written as OP square minus OT square as OA is equal to OT which is the radius of the circle OA and OT. So in place of OA we can write OT and OP square minus OT square will give us sorry OT square will give us PT square right and this will be by again Pythagoras theorem. Right so we can say that therefore A into PB is equal to PT square and this we have proved. This completes the first part of the solution. Now in the second part to prove E is equal to RC. Let me repeat the question once again here we are given that two circles are intersecting at P and Q from a point R on PB which is produced two tangents RB and RC are drawn to the two circles touching at B and C respectively. We need to prove that RB is equal to RC. So let us proceed with our proof. Here we are given that RB is the tangent to the circle is a C ket right. So using the above theorem which we have proved just now we can say that RB square is equal to RQ into RP that is proved by the given theorem. Now similarly RC is also tangent so RC square will be equal to RQ into RP. This is again by using the above term which we have proved just now. So from one and two from this equation and this equation we get that RB square is equal to RC square since right in size of both these equations are same to each other. So when square of two are equal then we can say that RB is equal to R, C and hence we have proved the second part also. So hope you understood the whole question well and using the above theorem which we have proved just now we can prove the second part easily. So have a nice day ahead.