 Hi friends, I am Purvan. Today we will discuss the following question. Using integration, find the area of the region enclosed between the circles x square plus y square is equal to 1 and x minus 1 whole square plus y square is equal to 1. Let us begin with the solution now. Now the equation of the circle is of the form x minus a whole square plus y minus b whole square is equal to r square where a, b are the coordinates of the center of the circle and r is the radius of the circle. Now here the equations of the circles are x square plus y square is equal to 1 and x minus 1 whole square plus y square is equal to 1. Now comparing this equation, x square plus y square is equal to 1 and the equation x minus a whole square plus y minus b whole square is equal to r square. We can clearly see that here radius is equal to 1 and the coordinates of the center of the circle are 0, 0. And comparing the equation x minus 1 whole square plus y square is equal to 1 with this equation x minus a whole square plus y minus b whole square is equal to r square. We can clearly see that here the coordinates of the center of the circle are 1, 0 and the radius of the circle is 1. Now we represent these both circles on the graph. Now here this is the circle whose equation is x square plus y square equal to 1. Now here we can clearly see that the radius of the circle is equal to 1 unit and the coordinates of the center of the circle are 0, 0 and this is the circle whose equation is x minus 1 whole square plus y square equal to 1. Here again the radius of the circle is 1 unit and the coordinates of the center of the circle are 1, 0. Now here we mark this as equation 1 and we mark this as equation 2. So 1 implies y square is equal to 1 minus x square. Now putting this value of y square in this equation 2 we get x minus 1 whole square plus 1 minus x square is equal to 1. Or we can write this as now x minus 1 whole square can be written as x square minus 2x plus 1 since we know that a minus b whole square is equal to a square minus 2ab plus b square. So applying this formula here we get x minus 1 whole square is equal to x square minus 2x plus 1 plus 1 minus x square is equal to 1. Now here x square, x square cancels out. So we get minus 2x plus 2 is equal to 1 or we can write this as minus 2x plus 1 is equal to 0. Therefore we get x is equal to 1 upon 2. Now here in this figure this shaded portion is the region enclosed between the two circles and also the figure is symmetric about x axis. So the area of this shaded portion is given by now since this figure is symmetric about x axis so if we find the area of oacda and multiply it by 2 we will get the area of this shaded portion. Now area of oacda is equal to area of oado plus area of dacd. So we have the area of the shaded portion is equal to 2 into area of oado plus area of dacd. This is equal to 2 into integral of limit from 0 to 1 by 2 as this point d has coordinates 1 by 2, 0. And this circle has equation x minus 1 whole square plus y square is equal to 1. Therefore we have area of oado is equal to integral of limit from 0 to 1 by 2 under root of 1 minus x minus 1 whole square dx. Now the equation of this circle is x square plus y square equal to 1 so we get y is equal to under root of 1 minus x square and here the limit is from 1 by 2 to 1. So we get plus now area of dacd is equal to integral of limit from 1 by 2 to 1 under root of 1 minus x square dx. Now this is equal to 2 into now integrating under root of 1 minus x minus 1 whole square we get x minus 1 upon 2 into under root of 1 minus x minus 1 whole square plus 1 upon 2 sin inverse x minus 1 and we have limit is from 0 to 1 by 2. This is since we know that integral of under root a square minus x square is equal to 1 upon 2 into x into under root of a square minus x square plus a square by 2 sin inverse x upon a plus c. So applying this formula here we get x minus 1 upon 2 into under root of 1 minus x minus 1 whole square plus 1 upon 2 sin inverse x minus 1. Here we can clearly see that a is 1 and x is x minus 1 plus now again applying this formula here we get x upon 2 into under root of 1 minus x square plus 1 upon 2 into sin inverse x and here limit is from 1 by 2 to 1. This is further equal to 2 into now putting the limits here we get putting up a limit 1 by 2 in place of x we get 1 by 2 minus 1 upon 2 into under root of 1 minus 1 by 2 minus 1 whole square plus 1 by 2 sin inverse 1 by 2 minus 1 minus putting lower limit that is 0 we get 0 minus 1 by 2 into under root of 1 minus 0 minus 1 whole square plus 1 by 2 sin inverse 0 minus 1. Now here putting the limits we get first we put up a limit 1 in place of x and we get plus 1 by 2 into under root of 1 minus 1 square plus 1 by 2 sin inverse 1 minus putting lower limit 1 by 2 in place of x we get 1 by 2 upon 2 into under root of 1 minus 1 by 2 whole square plus 1 by 2 into sin inverse 1 by 2. This is equal to 2 into now solving 1 by 2 minus 1 upon 2 we get minus 1 by 4 into under root of 1 minus now 1 by 2 minus 1 gives minus 1 by 2 and minus 1 by 2 whole square is equal to 1 upon 4 so we get 1 minus 1 by 4 plus 1 by 2 into sin inverse 1 by 2 minus 1 gives minus 1 by 2 minus 0 minus 1 by 2 gives minus 1 by 2 under root of 1 minus now 0 minus 1 whole square is equal to 1 square that is 1 so we get under root of 1 minus 1 plus 1 upon 2 into sin inverse minus 1 plus 1 by 2 into now under root of 1 minus 1 square is equal to 0 so we get 1 by 2 into 0 plus 1 by 2 into sin inverse 1 minus now 1 by 2 upon 2 is equal to 1 upon 4 so we get 1 upon 4 into under root of 1 minus now 1 by 2 whole square is equal to 1 by 4 plus 1 by 2 into sin inverse 1 by 2. This is equal to 2 into minus 1 by 4 into now under root of 1 minus 1 by 4 is equal to under root of 3 by 2 plus 1 by 2 into now sin inverse minus 1 by 2 is equal to minus pi by 6 so we get 1 by 2 into minus pi by 6 minus now under root of 1 minus 1 is equal to 0 so we get minus 0 minus 1 by 2 into now sin inverse of minus 1 is equal to minus pi by 2 so we get minus 1 by 2 into minus pi by 2 plus now 1 by 2 into 0 is 0 plus 1 by 2 now sin inverse 1 is pi by 2 so we get 1 by 2 into pi by 2 minus 1 by 4 into now under root of 1 minus 1 by 4 is equal to under root of 3 by 2 so we get 1 by 4 into root 3 by 2 plus 1 by 2 now sin inverse 1 by 2 is equal to minus pi by 6 so we get 1 by 2 into minus pi by 6 this is further equal to 2 into minus root 3 by 8 minus pi by 12 plus pi by 4 plus pi by 4 minus root 3 by 8 minus pi by 12 this is equal to 2 into now solving this bracket we get pi by 3 minus root 3 by 4 which is further equal to 2 pi by 3 minus root 3 by 2 square units so we have got our answer as 2 pi by 3 minus root 3 by 2 square units hope you have understood the solution and thank you