 Hello and welcome to the session. Let's discuss the following question. It says formula differential equations of the family of circles with their centers on the x-axis. So, let's now move on to the solution. Now, we know that if the center of the circle lies on x-axis, then y-coordinate must be 0. So, suppose the center of the circle is a 0 8. Now, the family of circles having center on the x-axis is given by x-a whole square plus y-0 whole square is equal to a square. This is how we give the equation of the circle and here a is arbitrary constant. Now, we simplify this. So, this becomes x square plus a square minus 2ax plus y square is equal to a square. a square gets cancelled with a square. So, we have x square plus y square minus 2ax is equal to 0. Now, since this equation contains only one arbitrary constant to find the differential equation, we need to differentiate this only once. So, differentiating this with respect to x we have 2x plus 2y dy by dx minus 2a is equal to 0. So, this implies x plus y dy by dx minus a is equal to 0. So, this implies a is equal to x plus y dy by dx. Let us name this as 1. Now, put a is equal to x plus y dy by dx in 1. So, we have x square plus y square minus 2ax a is x plus y dy by dx into x is equal to 0. Again, this is equal to x square plus y square minus 2x square minus 2xy dy by dx is equal to 0. So, this implies y square minus x square is equal to 2xy dy by dx and this is the required differential equation. So, this completes the question and the session by for now take care and have a good day.