 Hello and welcome to the session. In this session we will discuss a question which says that if the length of the tangent from the point fg to the circle x square plus y square is equal to 6, d twice the length of the tangent from the same point to the circle x square plus y square plus 3x plus 3y is equal to 0, then prove that f square plus g square plus 4f plus 4g plus 2 is equal to 0. Now before starting the solution of this question we should know a result and that is the length of the tangent drawn from the point x1, y1 to the circle whose equation is x square plus y square plus 2gx plus 2y plus c is equal to 0 is given by square root of x1 square plus y1 square plus 2gx1 plus 2xy1 plus c. Now this result will work out as a key idea for solving out this question. And now we will start with the solution. First of all we will find the length of the tangent from the point fg to the circle. Now given the equation of circle as x square plus y square is equal to 6 which implies x square plus y square minus 6 is equal to 0. Now we can find the length of the tangent from a point to a circle by using this formula. Now let us name it as 1. So the length of the tangent from the point fg to the circle whose equation is given as 1 is given by square root of f square plus g square minus 6. Here in this equation we have substituted the value of x as f and the value of y as g. And now in the second case we have to find the length of the tangent from this point which is the same point to this circle. Also it is given the equation of another circle is square plus y square plus 3x plus 3y is equal to 0. Now let us name it as perpendicular tangent from the point fg to the circle whose equation is given as equation number 2 is given by square root of 9. Now here substitute the value of x as f and the value of y as g. So it will be f square plus g square plus 3f. Now according to the condition which is given in the equation the length of the tangent from this point to this circle is twice the length of the tangent from this point again to the second circle. According to the question we have square root of f square plus g square minus 6 is equal to 2 into square root of f square plus g square plus 3f plus 3g. Which means the length of the tangent from the point fg to the first circle is equal to twice the length of the tangent from the point fg to the second circle. Now squaring both sides plus g square minus 6 is equal to 4 into f square plus g square plus 3f plus 3g the whole. Which implies f square plus g square minus 6 is equal to 4f square plus 4g square plus 12f plus 12g. On solving this implies 3f square plus 3g square plus 12f plus 12g plus 6 is equal to 0. This implies taking 3 common written brackets it will be f square plus g square plus 4f plus 4g plus 2 is equal to 0. Further this implies g square plus 4f 4g plus 2 is equal to 0. So in the question we have to prove this condition and it's proved. So this is the solution of the given question and that's all for the session. Hope you all have enjoyed the session.