At which point the tangent to the curve `x^(2)+y^(2)=25` is parallel to the line `3x-4y=7`?

Options

A. `(3,4),(-3,-4),`
B. `(3,-4),(-3,4)`
C. `(4,3)(-4,-3)`
D. `(-4,3)(4,-3)`

Answer

B

Explanation

Given `x^(2)+y^(2)=25`…………….i
`implies2x+2y(dy)/(dx)=0implies(dy)/(dx)=(-x)/y`
Now, slope of the line `3x-4y=78` is `m=3/4`.
Since, tangent is parallel to given line.
`:.(dy)/(dx)=3/4implies-x/y=3/4`
`impliesy=-4/3x`..ii
From Eqs i and ii we get
`x^(2)+16/9x^(2)=25implies(25x^(2))/9=25`
`implies x=+-3`
If `x=3` then from Eq. (ii) `y=-4`
If `x=-3`, then from Eq. (ii)
`:.` Points of contact are `(3,-4)` and `(-3,4)`