P is a point on the circle `x^2+y^2=9` Q is a point on the line `7x+y+3=0`. The perpendicular bisector of PQ is `x-y+1=0`. Then the coordinates of P are:

Options

A. `(0,-3)`
B. `(0,3)`
C. `(72//25,21//25)`
D. `(-72//25,21//25)`

Answer

4

Explanation

Any point on the line `7x+y+3=0` is `Q(t ,-3-7t),t inR`.
Now, `P(h,k)` is the image of point Q in the line `x-y+1=0`.
Then,
`(h-t)/(1)=(k-(-3-7t))/(-1)`
`=-(2{t-(-3-7t)+1})/(1+1)`
`=-8t-4`
or `(h,k) -= (-7t-4,t+1)`
This point lies on the circle `x^(2)+y^(2)=9`. Then,
`(-7t-4)^(2)+(t+1)^(2)=9`
or `50t^(2)+58t+8=0`
or `25t^(2)+29t+4=0`
or `(25t+4)(t+1)=0`
or `t=(-4)/(25),t= -1`
i.e., `(h,k) -= (-(72)/(25),(21)/(25))` or `(3,0)`