 Hello and welcome to the session. Let's discuss the following question. It says draw a circle of radius 3 cm, take 2 points p and q on one of its extended diameter, each at a distance of 7 cm from its centre, draw tangents to the circle from these 2 points p and q. So let's now move on to the solution. And we'll be writing the steps of construction. The first step is take a point o with o as centre and radius 3 cm, draw a circle. So we have drawn a circle with centre o and radius 3 cm. The second step is draw x o y, a diameter of the circle. So we have drawn the diameter of the circle, the diameter is x o y. The next step is mark 2 points p and q each at a distance of 7 cm from the centre o. So we have marked the points p and q each at a distance 7 cm from the centre o. The next step is draw the right bisector o p and o q intersecting o p, o q and b respectively. So we have drawn the right bisector of o p and o q intersecting o p at a and o q at b. Now the next step is with o as centre and radius equal to o a, draw a circle intersecting the circle of radius of 3 cm l and m. We have to draw a circle with centre a and radius p a and radius is p a. So we have drawn a circle with a as centre and p a as radius and it intersects the circle of radius 3 cm at the points l and m. Now the next step is join p l and p m. Now the next step is with b as centre, radius b q draw a circle intersecting the circle of radius 3 cm c and d. So we have drawn a circle with b as centre and b q as radius and it intersects the circle of radius 3 cm at the points c and d. Now the next step is join c q and c d, join c q and d q. So we have joined c q and d q, then p l and p m are the required engines from CQ and DQ are the required tangents to the point Q. Now we will justify that these are tangents that is PL, PM are the tangents, CQ and DQ are the tangents. The justification of POQ is the extended diameter of the circle and also angle PLO and angle PMO equal to 90 degrees and similarly angle QCO and angle QDO is equal to 90 degrees since these are angles in a semicircle. We can very well see that PLO is an angle in the semicircle of this circle and similarly PMO is an angle in a semicircle of this circle and similarly QCO is again the angle in the semicircle of this circle and QDO is the angle in the semicircle of this circle so these are 90 degrees each and also OLOM OC and OD are the diameters of the given circle that is the circle of radius 3 centimeter the given circle therefore OL is perpendicular to PLOM is perpendicular to PM OC is perpendicular to QC and OD is perpendicular to QD so OL is perpendicular to PLOM is perpendicular to PM OC is perpendicular to QC and OD is perpendicular to DQ so we can say that PL, PM, QC and QD are the tangents as we know by the theorem that tangent at any point of the circle is perpendicular to the radius through the point of contact so the radius is perpendicular to this line so all these are tangents so we can say hence PL, PM and QC, QD are the tangents to the given circle so this completes the question and the session bye for now take care have a good day