Since ` x ^(2) + y ^(2) = 1 ` a circle, `S_1` has centre `(0, 0)` and the X -axis at ` P (-1, 0) and Q (1, 0)`. Now, suppose the circle `S_2 `, with centre at `Q (1, 0)`has radius r. Since, the circle has to meet the first cricle , `0 lt r lt 2 `
Again, equation of the circle with centre at `Q (1, 0)` and radius r is
`" " (x - 1)^(2) + y^(2) = r ^(2)`
To find the coordinates of point R, we have to solve it with
`" " x^(2) + y ^(2) =1 `
On subtracting Eq. (ii) from Eq. (i), we get
`" " ( x -1 ) ^(2) - x ^(2) = r ^(2) - 1 `
`rArr x ^(2) + 1 - 2x - x ^(2) = r ^(2) -1 `
`rArr " " 1 - 2x = r ^(2) - 1 `
` therefore" " x = ( 2- r ^(2))/( 2)`
On putting the value of `x` in Eq. (i), we get
`" " (( 2-r ^(2))/( 2))^(2) + y ^(2) = 1 `
`rArr " " y ^(2) = 1 - (( 2 - r^(2))/( 2)) ^(2) = 1 - (( 2- r ^(2))^(2)))/( 4)`
`" " = 1 - ( r ^(4) - 4 r ^(2) +4 )/(4)`
` " " = ( 4 - r ^(4) + 4r^( 2 ) - 4 )/( 4)`
`" " = ( 4 r^(2) - r ^(4))/( 4)`
`" " = ( r^(2) ( 4- r ^(2))/(4) `
`rArr " " y = (r sqrt( 4- r ))/( 2)`
Again we know that, coordinates of S are ` (1- r, 0)`, therefore,
`" " SQ = 1 - (1- r) = r `
Let A denotes the area of `Delta QSR`, then
` " " A = (1)/(2) r [ r (sqrt ( 4- r ^(2))) /(2)]`
`" " = (1)/(4) r ^(2)sqrt ( 4 - r ^(2))`
` rArr " " A ^(2) = (1)/( 16) r^(4) ( 4- r ^(2))`
Let ` " " f ( r) = r ^(4) (4- r^(2)) = 4r ^(4) - r ^(6)`
`rArr " " f ' (r) = 16 r ^(3) - 6 r ^(5) = 2r^(3) ( 8 - 3r^(2))`
For maxima and minima, put ` f' (r) = 0 `
`rArr " " 2r ^(3) ( 8 - 3 r^(2)) = 0 `
`rArr " " r = 0, 8 - 3r ^(2) = 0`
`rArr " " r = 0, 3r^(2) = 8`
`rArr " " r = 0, r^(2) = 8//3`
`rArr " "r= 0 , r = ( 2 sqrt 2)/( sqrt 3)`
`" " [ because 0 lt r lt 2," so " r = 2 sqrt //sqrt 3]`
Again ` " " f'' (x) (r) = 48 r^(2) - 30 r ^(4)`
`rArr " " f'' ((2 sqrt 2)/(sqrt3))= 48 (( 4xx 2)/( 3)) - 30 (( 4xx 2 ) /(3))^(2)`
`" " = 1 6 xx8 - ( 10 xx 64)/( 3) = 128 - ( 640)/(3) = - ( 256)/( 2) lt 0`
Therefore, `f (r)` is maximum when, ` r = (2 sqrt 2)/( sqrt 3)`
Hence, maximum value of A
`" " = (1)/(4) ((2 sqrt 2)/(sqrt3))^(2) sqrt ( 4 - (( 2sqrt 2)/( sqrt3))^(2)) = (1)/(4) ((8)/(3)) * sqrt ( 4 - (8)/(3)) `
` " " =(2)/(3) * ( sqrt ( 12 - 8))/( sqrt 3) = ( 2* 2)/(3sqrt 3) = ( 4)/(3sqrt 3) = ( 4sqrt 3)/(9)`
