 Hello everyone, Myself Professor Pirti Stutte working as an assistant professor in mechanical engineering department as an in Volchian Institute of Technology, Singapore. So in this session we will learn how to derive the equation for the circle that is also called as parametric representation of a circle which is the part of the course CADMCA. So what you are learning you can apply the CAD fundamentals like the geometrical transformation or may be for example, how to apply the applications for the geometrical modeling which is the part of the parametric representation of a line or may be the circle. So what we are discussing today in this session first of all we will see the recap of the last session in that we have discussed the parametric representation of a line. In this session we will discuss the parametric representation of a circle, the derivation and the based on the derivation we will see the numerical very firstly we will go through the quick review of the parametric representation of a line. So which consists of the start point of the line and the end point of the line where u is equal to 0 and u is equal to 1 at the end point where u is the parameter p1 and the p2 are the position vector at the start point and the end point of the line. So in general parametric equation of this line is the p1 plus p2 minus p1 of u where p1 and p2 are the position vector and u is the parameter. You need vector that is p2 minus p1 divided by L where L is the length of the line. We can also calculate the length of the line with the help of the coordinates like x1, y1, z1 and x2, y2, z2 these are the coordinates of the start point and the end point of the line respectively where length of the line that will be under root x2 minus x1 bracket square plus y2 minus y1 bracket square plus z2 minus z1 bracket square. In the Cartesian form suppose I want to write the parametric equation of a line with respect to x y z that is x of u is equal to x1 plus x2 minus x1 of u. The in general equation of a line that will be p of u is equal to p1 plus p2 minus p1 into u. This is the very similar equation with respect to the Cartesian form. y of u is equal to y1 plus y2 minus y1 of u and z of u is equal to z1 plus z2 minus z1 of u. You can compare these equations that is in the Cartesian form and the in general form you can compare these are very similar. Again we will be needing for example the parametric equation of a line for example we will keep the constraint values that is u is equal to 0 and u is equal to 1 in the in general parametric equation of a line. For example we will keep u is equal to 0 so what we will have it so p of u is equal to 0. So p1 plus p2 minus p1 into 0 that will be p1 so at u is equal to 0 that will be p1. So at the end point that is the another constraint that is u is equal to 1 provide u is equal to 1 here that is p1 plus p2 minus p1 p1 will be deleted so that will be p2 at u is equal to 1. So the we can call this the derivation with respect to you regarding this equation that is d by du of p1 plus p2 minus p1 into u. So the remaining part will be the derivation that will be p2 minus p1. So p dash of u is equal to p2 minus p1 after that. So similarly we can write it in the Cartesian form that is x of u is equal to x2 minus x1 y dash of u is equal to y2 minus y1 and z dash of u is equal to z2 minus z1. The slope of the curve that is y dash divided by x dash that is dy by dx y dash here you can write it as y2 minus y1 and x dash you can write it as x2 minus x1. So we will see one the numerical based on the parametric equation of a line. We have to determine the parametric representation of a line of the segment between the position vector that is p1 and p2 that is x1 y1 and x2 y2, x1 is 1, y1 is 1, x2 is 4, y2 is 4, 5. Also we have to determine the tangent vector and the slope of the curve that is a line. So first we will derive the parametric equation of the line in the Cartesian form. So x of u is equal to x1 plus x2 minus x1 of u. So that will be x of u is equal to 1 plus 4 minus 1. So x1 is 4, x2 is 4, x1 is 1. So that will be 1 plus 3 u after that y of u is equal to y1 plus y2 minus y1 of u. So by keeping the values regarding y1 and y2, so y1 is 1, y2 is 5 and y1 is 1. So that will be 1 plus 4 u after that in the in general equation form. So general parametric equation of a line that will be p1 plus p2 minus p1 of u. So p1 is 1, 1, p2 is 4, 5. So that will be 1, 1 plus 3, 4 of u. Tangent vector that is p2 minus p1. So here we are having the p2 minus p1 that is 3, 4 of u that will be for example the tangent vector. The slope of the curve that will be y dash divided by x dash. So here y dash divided by x dash that is 4 divided by 3 that is 1.33 that will be the slope of the curve. So these are the some of the we can call the parametric equation of the line, the derivation and the numericals we have discussed it in the last session. Now so today we will discuss the parametric equation of a circle. So the circle consists of the radius r, yes the point of p xy which is located on the circle which is with respect to angle theta, with respect to the radius r along with the axis x, yes now. So r and theta will be needed and x and y that will be needed because p xy that is a point on the circle. So implicit form of the circle that is r square plus x square plus y square that is r is the radius and x and y are the point which is which are located on the circle. So r square is equal to x square plus y square that is the equation for the circle. So r is equal to under root x square plus y square. So the equation of the circle, another equation of the circle that is where the center of the circle that will be a and b here in this point the center of the circle is 0 0. So if suppose the center of the circle is a comma b, so what will happen? So that will be r is equal to x minus a bracket square plus y minus b bracket square. So here it will be suppose your center is at the 0 0, so x minus 0 bracket square plus y minus b bracket minus 0 bracket square that will be x square plus y square regarding r. But here another form that is if suppose the center of the circle is a comma b. So in that case, so r is equal to under root x minus a bracket square plus y minus b bracket square. So parametric equation of the circle, so in general equation suppose we want to discuss we will need that is x is equal to r cos theta. So x is equal to r cos theta and y is equal to r sin theta, yes so y is equal to r sin theta. So with the help of this in generally we can called as the parametric equation of the circle that is x of u is equal to r into y minus u bracket square plus y 1 plus u square and y of u is equal to r into 2 u divided by 1 plus u square. So this is the in general parametric equation of the circle where x is equal to r cos theta and y is equal to r sin theta. So based on this we will discuss the equation. Now represent the circle with the center 3 comma 4 and the radius 50 mm construct the implicit and the parametric equation. Now first we will discuss the equation of the circle in the implicit form. So r is r square is equal to x square plus y square whenever the center is 0 0. But here there is a center 3 comma 4 that is provided. So that will be r square is equal to x minus a bracket square plus y minus b bracket square after that. So r is equal to in the bracket x minus a bracket square plus y minus b bracket square. So 50 is equal to x minus 3 bracket square plus y minus 4 bracket square. This is the parametric we can called as a parametric equation of the circle in implicit form. So in general parametric equation of the circle so x of u is equal to r into 1 minus u square plus 1 minus u square y of u is equal to r into 2 u divided by 1 plus u square by keeping the values regarding the r that is 50 into 1 minus u square plus 1 plus u square and y of u is equal to 50 into 2 u divided by 1 plus u square. So these are the very simple numerical based on the parametric equation of the circle based on the implicit form and the in general form. Now you are having one question that is what is the difference between analytical and synthetic curve you can think about this yes. So analytical curve that we have discussed like may be the circle or may be the line ellipse hyperbola but synthetic curve that are the very much important part of cadkem say like may be the B spline or may be the Hermite cubic spline or may be the Bezier curve these are the synthetic curve. These are the references thank you.