 Hello and welcome to the session. In this session first we are going to discuss equation of the tangent at a point on the circle. As we know that the tangent at a point to a circle is perpendicular to the radius of the circle through the point of contact. Using this concept we find the equation of the tangent. First find the slope of the radius through the point of contact as tangent is perpendicular to the radius. The slopes are negative reciprocals of each other by using the point slope form that is y minus y1 is equal to m into x minus x1. We can find the equation of the tangent where point x1 y1 is the point of contact of tangent to the circle. Let us consider an example. Find the equation of the tangent to the circle x square plus y square plus x minus y minus 2 is equal to 0 at the point p with the coordinates 1 1. We have given the equation of the circle as x square plus y square plus x minus y minus 2 is equal to 0 and the general equation of the circle is given by x square plus y square plus 2 gx plus 2 at y plus c is equal to 0. On comparing the two equations we get the value of 2g is equal to 1 which implies that g is equal to 1 by 2 and the value of 2 earth is equal to minus 1 which implies that earth is equal to minus of 1 by 2. And we know that the center of the circle is given by minus of g minus circle so we have the center of the given circle as minus of 1 by 2 1 by 2. The center of the given circle is given by minus 1 by 2 1 by 2 and we need to find the equation of the tangent to the circle at the point p with coordinates r 1 1. Let the center of the circle is denoted by the point o with the coordinates minus 1 by 2 1 by 2. Then slope of o p is given by y 2 minus y 1 upon x 2 minus x 1 that is 1 minus of 1 by 2 upon 1 minus of minus 1 by 2 which is equal to 1 by 2 upon 3 by 2 that is 1 by 3 as the tangent at p is perpendicular to the radius o p. Therefore slope of tangent which is equal to negative reciprocal of the slope of radius o p is given by minus 3 therefore equation of the tangent to the given circle at the point y 1 is given by y minus 1 is equal to minus 3 into x minus 1 which can be written as y minus 1 is equal to minus of 3 x plus 3 that is we have y plus 3 x minus 1 minus 3 is equal to 0. But this can be written as y plus 3 x minus 4 is equal to 0 which is the required equation. Let us now discuss equation of a tangent to the circle x square plus y square is equal to r square in terms of its gradient. Let the equation of the circle be x square plus y square is equal to r square and mark this equation as 1. Let the equation of the tangent to the circle is given by y is equal to n into x plus c and mark this equation as 2. Now solving equation 1 and equation 2 simultaneously we get x square plus y square and y is given by n into x plus c so we can write n into x plus c the whole square is equal to r square. Therefore we have x square plus n square into x square plus c square plus twice of n into x into c is equal to r square. And we can also write it as taking x square common from the first two terms we get x square into 1 plus n square plus twice of n into x into c plus c square minus of r square is equal to 0. Mark this equation as 3. The equation y is equal to n into x plus c will be tangent to the circle x square plus y square is equal to r square. The above equation that is equation number 3 has equal roots that is v square minus 4 into a into c is equal to 0 as discriminant will be 0. Therefore we have v square that is square of the square to the root of x which is given by 2 into n into c the whole square minus of 4 into a that is the square to the root of x square given by 1 plus n square into c that is c square minus r square is equal to 0. Therefore we have 4 into n square into c square minus of 4 into c square plus c square into n square minus r square minus of r square into n square is equal to 0. Which implies that 4 into n square into c square minus of 4 into c square into n square minus of 4 into c square minus r square minus r square into n square is equal to 0. Which implies that c square is equal to r square into 1 plus m square or we can say c is given by plus minus of r into square root of 1 plus m square. My letter substitute the value of c when y is equal to m into x plus c then we will get the equation of the tangent to the circle x square plus y square is equal to r square. We have y is equal to m into x plus c therefore we can write it as y is equal to m into x plus c and c can be written as plus minus r into square root of 1 plus m square. So we have plus minus r into square root of 1 plus m square where m can take any value which implies that y is equal to m into x plus r into square root of 1 plus m square or y is equal to m into x minus of r into square root of 1 plus m square. If the equation of the tangent d equation y is equal to m into x plus c will be a tangent to the circle x square plus y square is equal to r square if either c is equal to plus r into square root of 1 plus m square or c is equal to minus of r into square root of 1 plus m square which is the condition of the tangent. The reason why there are two equations that is y is equal to mx plus minus of r into square root of 1 plus m square is that there are always two tangents with the given slope to the circle namely the tangents at the end of the diameter. We are now going to discuss how to find the equation of the circle through the intersection of the given circle and the given line. Let s be identical equal to x square plus y square plus twice of d into x plus twice of f into y plus c is equal to 0 be the equation of a circle and let l be identical equal to l into x plus m into y plus m is equal to 0 be the equation of a line. The equation s plus lambda l is equal to 0 that is x square plus y square plus twice of d into x plus twice of f into y plus c plus lambda into l into x plus m into y plus m is equal to 0. Represent the equation of the circle through the intersection of the circle and the line. Let us take an example the equation of the curve of the circle x square plus y square is equal to r square is mx plus my plus m is equal to 0 find the equation of the circle drawn on the curve at the diameter. Now we are given a circle x square plus y square is equal to r square with center o and the curve of the circle is given by the equation mx plus my plus m is equal to 0. We need to find the equation of the circle drawn on the curve at diameter that is we need to find the equation of the new circle with center o dash and a b of diameter. Therefore equation of any circle through the intersection of x square plus y square is equal to r square and mx plus my plus m is equal to 0 is given by x square plus y square minus of r square plus lambda into delta. x plus n y plus n is equal to 0 which can be written as x square plus y square plus lambda times lx plus lambda into ny plus lambda n minus of r square is equal to 0. Now we have the equation of the circle as x square plus y square plus lambda times l into x plus lambda into n into y plus lambda n minus r square is equal to 0 and we know that the general equation of the circle is given by x square plus y square plus twice of g into x plus twice of f into y plus v is equal to 0 which center minus g minus f. Therefore the center of this circle will be equal to minus of lambda into l by 2 minus of lambda into n by 2. The center of this circle is equal to minus of lambda into l by 2 minus of lambda into n by 2 as a b is the diameter of the new circle. Therefore the center that is minus lambda into l by 2 minus lambda into n by 2 will lie on lx plus ny plus n is equal to 0 that is the equation of the diameter a b which implies that l into x that is minus lambda into l by 2 plus n into y that is n into minus lambda into n by 2 plus m is equal to 0 which implies that minus of lambda into l square by 2 minus of lambda into n square by 2 plus m is equal to 0. Now taking minus lambda by 2 commons from the first two terms we get minus lambda by 2 into l square plus n square plus m is equal to 0 which implies that minus lambda by 2 into l square plus n square is equal to minus of m. Therefore we get minus lambda into l square plus n square is equal to minus of twice of m that is lambda into l square plus m square is equal to twice of m which implies that the value of lambda is equal to twice of m upon l square plus m square. Now putting the value of lambda in this equation we get the required equation of the new circle therefore the required equation is x square plus y square minus of r square plus lambda that is twice of m upon l square plus m square into l into x plus m into y plus twice of m square is equal to 0 and solving this equation we get n square plus n square into x square plus n square plus n square into y square plus twice of m into l into x plus m into m into y plus twice of n square minus r square into a square plus n square is equal to 0 which will be required equation of the new circle with a b as biometer. This completes our session. Hope you enjoyed this session.