 Hi friends I am Purva and today we will discuss the following question. Choose the correct answer in the following and the question is the area of the circle x square plus y square is equal to 16. Exterior to the parabola y square is equal to 6x is a 4 upon 3 into 4 pi minus root 3 b 4 upon 3 into 4 pi plus root 3 c 4 upon 3 into 8 pi minus root 3 d 4 upon 3 into 8 pi plus root 3. Let us begin with the solution now. Now first we will find the points of intersection of the circle x square plus y square is equal to 16 and the parabola y square is equal to 6x. Now let us mark this as equation 1 and we mark this as equation 2. Now from equation 1 and 2 we get. Now we substitute the value of y square is equal to 6x from equation 2 into equation 1. So we get x square plus 6x is equal to 16 and this implies x square plus 6x minus 16 is equal to 0 and this further implies x square plus 8x minus 2x minus 16 is equal to 0 and this implies now taking out x common from first two terms we get x into x plus 8 and taking out minus 2 common from the next two terms we get minus 2 into x plus 8 and this is equal to 0 and this implies x plus 8 into x minus 2 is equal to 0 and this implies either x is equal to minus 8 or x is equal to 2. Now if x is equal to minus 8 then putting the value of x in equation 2 we get y square is equal to 6 into minus 8 and this is equal to minus 48 which is not possible and if x is equal to 2 then putting this value in equation 2 we get y square is equal to 6 into 2 and this is equal to 12. So we get y is equal to plus minus 2 root 3. Therefore the points of intersection of the two curves are 2 comma 2 root 3 and 2 comma minus 2 root 3. Now we also know that x square plus y square is equal to 16 is a circle with center 0 0 and radius 4 so with this information we will draw a sketch to determine the region whose area is to be found out. So this is the parabola y square is equal to 6x and this is the circle x square plus y square is equal to 16 with radius 4 and these two points C and E are the points of intersection of the parabola and the circle and this shaded region is the region whose area is to be found out. Now this shaded region is symmetric about x axis therefore we have required area is equal to twice into area of A B D C M O A minus area of C M O C therefore we have required area is equal to 2 into area of A B D C M O A minus area of C O M C. Now area of A B D C M O A is bounded by the circle x square plus y square is equal to 16 and we have limit is from minus 4 to 2 therefore we get this is equal to 2 into now area of A B D C M O A is given by integral limit from minus 4 to 2 and equation of circle is x square plus y square is equal to 16 so we get y is equal to under root of 16 minus x square dx minus now area of C M O C is bounded by this parabola y square is equal to 6x and this line and limit is from 0 to 2 so we have minus integral limit is from 0 to 2 and equation of parabola is given by y square is equal to 6x so we have y is equal to under root 6x dx this is equal to 2 into now integral of under root 16 minus x square is given by x upon 2 into under root of 16 minus x square plus 16 upon 2 into sin inverse x upon 4 and we have limit is from minus 4 to 2 now we know that integral under root of a square minus x square dx is equal to x upon 2 into under root of a square minus x square plus a square upon 2 into sin inverse x upon 8 so here taking a is equal to 4 and x is equal to x we get integral of under root of 16 minus x square is equal to this minus now integral of root 6x is root 6 into x raise to the power 3 upon 2 divided by 3 upon 2 and here we have limit is from 0 to 2 and this is equal to 2 into now putting the limits we get first we put up a limit 2 in place of x and we get 2 upon 2 into under root of 16 minus 4 plus 8 into sin inverse 2 upon 4 minus now putting lower limit minus 4 in place of x we get minus 4 upon 2 into 0 plus 8 sin inverse minus 4 upon 4 minus now here putting up a limit 2 in place of x we get root 16 into 2 upon 3 into 2 raise to the power 3 upon 2 minus putting lower limit 0 in place of x we get 0 and this is equal to 2 into now canceling out 2 in numerator and denominator here we get under root of 12 plus 8 into now canceling out 2 here in numerator and denominator we get sin inverse 1 upon 2 and sin inverse 1 upon 2 is equal to pi by 6 so we get 8 into pi by 6 minus now this thing will become 0 so we get 8 into now canceling out 4 in numerator and denominator here we get sin inverse minus 1 and sin inverse minus 1 is equal to minus pi by 2 so we get minus 8 into minus pi by 2 minus now solving this bracket we get root 2 into root 3 into 4 upon 3 into 2 raise to the power 1 upon 2 this is equal to 2 into now we can write root 12 as 2 root 3 plus now canceling out common factor 2 from numerator and denominator we get 4 in numerator and 3 in denominator so we get 4 pi by 3 plus again canceling out 2 in numerator and denominator we get 4 in numerator so we get plus 4 pi because here minus into minus will become plus minus now solving this whole thing we will get 8 root 3 upon 3 and this is equal to 2 into 16 pi by 3 minus 2 root 3 upon 3 and this is further equal to 2 into 2 upon 3 into 8 pi minus root 3 we take out 2 upon 3 common from this bracket and we get 2 into 2 upon 3 into 8 pi minus root 3 and this is further equal to 4 upon 3 into 8 pi minus root 3 therefore we get the required area is equal to 4 upon 3 into 8 pi minus root 3 and this is same as option C hence we get the correct answer is C hope you have understood the solution by antique care