 Hello and welcome to the session. In this session we will discuss a question which says that find the radius and center of the circle z into the conjugate of z minus 3 plus 4 iota the whole into z minus 3 minus 4 iota the whole into the conjugate of z plus 16 is equal to 0 where z is a complex variable. Now before starting the solution of this question we should know a result and that is an imperative property of modulus. That is for a complex number z which is equal to a plus b iota z into the conjugate of z is equal to modulus of z square where the modulus of z is equal to square root of a square plus p square and the conjugate of z is 0 minus b iota. Now this result will work out as a key idea. Let's try to write this question and now we will start with the solution. We have to find the radius and the center of the circle where z is a complex variable. Now let z is equal to a plus b iota. Now using this result the conjugate of z is equal to a minus b iota. Also the modulus of z is equal to square root of a square plus p square. Therefore using this result which is given as a key idea, z into the conjugate of z is equal to modulus of z square which implies z into the conjugate of z is equal to square root of a square plus b square whole square which implies z into the conjugate of z is equal to a square plus p square. Now given the circle z into the conjugate of z plus 4 iota the whole into z minus 3 minus 4 iota the whole into the conjugate of z plus 16 is equal to 0. This is the conjugate of z z into the conjugate of z. Now putting all these values here plus b square minus iota the whole plus b iota the whole minus 3 minus 4 iota the whole into a minus b iota the whole is equal to 0. Now this implies a square plus b square the whole minus of 3 a minus 3 b iota minus 4 a iota plus 4 b iota square the whole plus b square minus 3 a minus 3 b iota minus 4 a iota minus 4 b into iota square is minus 1 minus 3 a plus 3 b iota iota minus 4 b iota square which is minus 1 plus 16 is equal to 0. Further this implies a square plus b square minus this will be cancelled with each other and here minus into minus will be plus where it will be plus 4 b minus 3 a and here also it is plus 4 b plus 16 is equal to 0. Now this implies plus b square and this will be minus 6 a and this will be plus 8 b plus 16 is equal to 0. Now this is the general equation of the circle where the sentence is given by this. Now comparing the equation of the circle here equal to minus 6 presenting the circle still is given by minus f which will be equal to minus 3 and minus which is equal to 3 minus 4 and the range of this circle which will be equal to square root of g square plus f square minus c. Now here it will be minus 3 square plus 4 square which is equal to square root of 9 plus 16 minus 16 which is equal to root 9 and which is equal to 3. Question number one, the circle which is given by equation number one. The equation one, the solution of the given question and that's all for this session. Hope you all have enjoyed this session.