 Hello and welcome to this session. In this session first we will discuss construction of circle-circle. Before we move on to the construction, first of all let's see what the circle-circle is. The point where the perpendicular bisectors triangle. Next we have the circle passing through the vertices of the triangle, the circle-circle. At a circle-circle for a given triangle, suppose the triangle PQR with PQ equal to 5 centimeters, 5 centimeters is of measure 50 degrees. The figure of the triangle PQR, this is the figure of measure 5 centimeters, PR is of measure 4 centimeters and angle QPR is of measure 50 degrees. First of all let's consider this to draw 2 of measure 5 centimeters. 2 Px of measure 50 degrees are where PR is of measure 4 centimeters to 4 centimeters. This point of intersection of the arc with Px is point and is of measure 4 centimeters. We get the required triangle PQR of which we need to construct the circle-circle. The circle-circle is the circle that circles the vertices of the given triangle. We know that the circle-center of the triangle is the point of intersection of the perpendicular bisectors of the size of the triangle. So, for the circle-circle of the triangle, first of all we need to circle-center of the perpendicular bisectors of the size of the triangle. Next step would be draw the perpendicular bisectors respectively. In the bisectors of the size PQ, let YZ and ST the perpendicular bisectors of size PQ and PR respectively intersect at the point of the triangle PQR. Now it may be taken as OP OQ or OR that is we can take the radius of the 3 vertices of the triangle PQR. When by taking OST center and OP ST radius we draw a circle in the PQR and radius equal to OP such that the circle passes through the points PQ and R of the triangle PQR. Circle-circle of the triangle PQR with OST circle-circle. The structure of the in-circle is in-center of the bisectors of the centimeters so equal to 4 centimeters. The PQR is of measure 5 centimeters, QR is of measure 3 centimeters and PR of measure 5 centimeters. Now from the rough figure we have of measures 4 centimeters and QR is also of measure 3 centimeters. So to locate the point that would be width and Q centimeters and these two arcs intersect at the point in the PQR in which PQ is of measure 5 centimeters and QR is of measure for which we have to construct the in-circle. Now since we know that in-circle is the biggest possible circle and whose center is the in-center should not be angle bisectors of the given triangle. First of all we need to find out the in-centers for this given first angle PQR the angle bisectors angle P and QY is the angle bisector of angle Q let them get intersect at the point. So now in the next step PQ is I here perpendicular to PQ the in-circle this is the biggest possible such that it touches the PQR the distance of the three sides of the triangle of the circum-circle and in-circle of the triangle.