 Hello and welcome to the session. Let's work out the following question. Let's say, drop a line segment A of 8 cm, taking A as center, draw a circle of radius 4 cm, taking B as center, draw another circle of radius 3 cm, construct pendants to 8 circles from the center of the other circle. So let's now move on to the solution, construction and we'll write the steps of construction side by side. First step is, draw a line segment of A of 8 cm. This is the line segment A of 8 cm. Now the next step is, with A as center and radius 4 cm, draw a circle. We have drawn a circle with A as center and radius 4 cm. Now the next step is, with and radius 3 cm, draw another circle. This is a circle with center B and radius 3 cm. Now the next step is, particular bisector of A of B, which intersects A of B at home. Now the next step is, center and radius equal to 2 O of B, draw a third circle and radius equal to O of A is equal to A of B at the point B and Q. Now A is the external point for the circle with center B and B is the external point for the circle with center A. Now the next step is, are the required tangents to the circle with center B. Similarly, join BP and BQ then and BQ are the tangents to the circle with center A. So we have joined AM, AM, BP and BQ and these are the required tangents. Now we will justify that these are tangents to the circle that is AM, AM, BP and BQ are the tangents. For that we will show that line drawn from the center of the circle to the tangents is perpendicular to the point of contact. As we know by theorem that tangent at any point of a circle is perpendicular to the radius through the point of contact is equal to AQB is equal to 90 degrees. And similarly angle BMA is equal to angle BNA is equal to 90 degrees. So angle APB is 90 degrees because it's an angle in semicircle of this circle. Similarly angle AQB is also 90 degrees that is this angle and this angle is 90 degrees because these are the angles in semicircle of this circle. So here we have APB not APQ, APB similarly angle BMA is 90 degrees and BNA is 90 degrees. Again these are the angles in semicircle so we will write the reason. Angle in a semicircle. Now since APB is 90 degrees, AQB is 90 degrees therefore AP is perpendicular to BP, APQ is perpendicular to BQ, BM is perpendicular to AM, BN is perpendicular to AN. Angle B is perpendicular to BP, AQ is perpendicular to VQ, BM is perpendicular to AM and BN is perpendicular to AN. That means the tangent at any point of the circle is perpendicular to the radius through the point of contact are the tangents to the circle with centre B are the tangents to the circle with centre A. So this completes the question and the session. Bye for now. Take care. Have a good day.