 Hi and welcome to the session. Let us discuss the following question. Question says, find the area bounded by the curves x-1 whole square plus y square is equal to 1 and x square plus y square is equal to 1. Let us now start with the solution. We are given equations of two circles. Equations are x-1 whole square plus y square is equal to 1 and x square plus y square is equal to 1. Now let us name this equation as 1 and this equation as 2. Now clearly we can see equation 1 is a circle with center 1, 0 and radius 1. Similarly equation 2 is a circle with center as origin and radius equal to 1. Now we can write equation 1 is a circle with center c that is 1, 0 and radius 1 and equation 2 is a circle with center o at origin and radius equal to 1. Now let us find out these two points that is points of intersection of the two circles. Now solving equations 1 and 2 we have x-1 whole square plus y square is equal to 1 whole square plus y square substituting x square plus y square for 1 here we get this equation. Now we will apply formula of a minus b whole square in this bracket. We know a minus b whole square is equal to a square plus b square minus 2ab. We will write plus y square as it is and here we will write x square plus y square. Now x square on left hand side and right hand side will get cancelled. Similarly y square on left hand side and right hand side will get cancelled. Now we get 1 minus 2x is equal to 0. Now this further implies minus 2x is equal to minus 1. Now we get x is equal to 1 upon 2 dividing both the sides of this expression by minus 2 we get x is equal to 1 upon 2. Now we know x square plus y square is equal to 1. Now y square is equal to 1 minus x square or we can say y is equal to square root of 1 minus x square. Now let us name this expression as 3. Now substituting value of x is equal to 1 upon 2 in expression 3 we get y is equal to square root of 1 minus square of 1 upon 2. Now this is further equal to square root of 4 minus 1 upon 4. Now this further implies y is equal to plus minus square root of 3 upon 2. Thus the points of intersection of the given circles are 1 upon 2 root 3 upon 2 and 1 upon 2 minus root 3 upon 2. Now this is the required area. Required area of the inflows region OACA dash O is equal to 2 multiplied by area of the region ODCAO. So we can write required area of the inflows region OACA dash O between circles is equal to 2 multiplied by area of the region ODCAO. Now this region ODCAO is equal to area of the region ODAO plus area of the region DCAD. So we can write area of the region ODCAO is equal to area of region ODAO plus area of region DCAD. Now area of this region that is ODAO is equal to integral from 0 to 1 upon 2 y dx. Now we will write plus sign as it is. Now area of the region DCAD is equal to definite integral from 1 upon 2 to 1 y dx. Now we know y of this region is equal to square root of 1 minus x minus 1 whole square. So we can write 2 multiplied by integral from 0 to 1 upon 2 square root of 1 minus x minus 1 whole square dx. We will write this plus sign as it is. Now y of this region is equal to square root of 1 minus x square. So we will write this integral as definite integral from 1 upon 2 to 1 square root of 1 minus x square dx. Now let us evaluate these two integrals. First of all let us recall formula of integration that integral of square root of a square minus x square dx is equal to 1 upon 2x multiplied by square root of a square minus x square plus a square upon 2 sin inverse x upon a plus c. Now we know 1 raised to any power is equal to 1 only. So here we can write 1 square. Similarly here also we can write 1 square. Now using this formula we get this integral is equal to 1 upon 2 multiplied by x minus 1 multiplied by square root of 1 square minus x minus 1 whole square plus 1 square upon 2 multiplied by sin inverse of x minus 1 upon 1 and limit of the integral varies from 0 to 1 upon 2. Now we will write this plus sign as it is. We know 2 will get multiplied by this integral as well as by this integral. So here we will write 2. Now let us find out this integral. Now this integral is equal to 1 upon 2 multiplied by x multiplied by 1 minus x square plus 1 upon 2 sin inverse x upon a and limit of this integral varies from 1 upon 2 to 1. Now substituting these limits in this function we get minus 1 upon 2 multiplied by square root of 1 minus 1 upon 4 plus sin inverse minus 1 upon 2 minus sin inverse minus 1. Now we will write this plus sign as it is. Now substituting these limits in this function we get sin inverse 1 minus 1 upon 2 multiplied by square root of 1 minus 1 upon 4 plus sin inverse 1 upon 2. Now simplifying this bracket we get minus root 3 upon 4. Simplifying this term we get minus root 3 upon 4. Now we know sin inverse minus 1 upon 2 is equal to minus pi upon 6. So here we will write minus pi upon 6. Now sin inverse minus 1 is equal to minus pi upon 2 and minus pi upon 2 multiplied by this minus sign is equal to plus pi upon 2. So here we will write plus pi upon 2. Now we will write this plus sign as it is. Now we will simplify this bracket. We know sin inverse 1 is equal to pi upon 2. So here we will write pi upon 2. Now simplifying this term we get root 3 upon 4 and root 3 upon 4 multiplied by this minus sign is equal to minus root 3 upon 4. Now we will write minus pi upon 6. We know sin inverse 1 upon 2 is equal to pi upon 6 and pi upon 6 multiplied by this minus sign is equal to minus pi upon 6. Now this expression is further equal to minus 2 root 3 upon 4 plus minus pi plus 3 pi plus 3 pi minus pi upon 6. Adding these two terms we get this term and adding rest of the terms by taking their LCM we get this term. Now this is further equal to minus root 3 upon 2 plus 4 pi upon 6. Now we will cancel common factor 2 from numerator and denominator both and we get 2 pi upon 3 minus root 3 upon 2. So we get area of the enclosed region OAC A-O between circles is equal to 2 pi upon 3 minus root 3 upon 2. So this is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.