 Good morning friends, I am Purva and today I will help you with the following question from the differential equation of the family of circles touching the y-axis at origin Let's begin with the solution now Now here we are required to find the family of circles touching y-axis at origin. So let's see denote the family of circles touching y-axis at origin and let A comma 0 be the coordinates of the center of any member of the family then the equation of family C is x minus a whole square plus y square is equal to a square Now as y-axis is a tangent. So the radius of the circle is equal to x Now this implies x minus a whole square can be written as x square minus 2a x plus a square plus y square is equal to a square Tantraling out a square from both the sides we get this further implies x square plus y square is equal to 2a x We mark this as equation one and here we have a is an arbitrary constant Now since there is only one constant, so we shall differentiate equation one only once to get the required differential equation so differentiating one with respect to x we get differentiating x square we get 2x Plus differentiating y square we get 2y into dy by dx is equal to differentiating 2a x we get 2a and we can write this as this implies x plus y is 2 now dy by dx is equal to y dash So we have y dash is equal to a and we mark this as equation two Now substituting the value of a from equation two In equation one we get x square plus y square is equal to 2x into x plus y into y dash This implies x square plus y square is equal to now opening the bracket we get 2x into x gives 2x square plus 2x into y into y dash gives 2x into y into y dash and this further implies y square is equal to now 2x square minus x square gives x square plus 2x into y into y dash Hence the required differential equation is 2x into y into y dash plus x square is equal to y square This is our answer. Hope you have understood the solution. Bye and take care