Find the equation of the tangents through (7,1) to the circle `x^2+y^2=25`

Options

A. `3x+4y-25=0, 4x-3y-25=0`
B. `4x+3y-31=0, 3x-4y-17=0`
C. `3x-2y-19=0, 2x+3y-17=0`
D. none of these

Answer

A

Explanation

The equation of any line through (7, 1) is
`y-1=m(x-7)rArrmx-y-7m+1=0 " " ...(i)`
The coordinates of the centre and radius of the given circle are (0, 0) and 5 respectively.
The line (i) will touch the given circle, if,
Length of the perpendicular from the centre = Radius
`rArr |(mxx0-0-7m+1)/(sqrt(m^(2)+(-1)^(2)))|=5`
`rArr |(1-7m)/(sqrt(m^(2))+1)|=5`
`rArr ((1-7m)^(2))/(m^(2)+1)=25rArr 24m^(2)-14m-24=0rArr m=-3//4, 4//3`
Substituting the values of m in (i), we obtain
`-(3)/(4)x-y+(21)/(4)+1=0` and `(4)/(3)x-y-(28)/(3)+1=0`
`rArr 3x+4y-25=0` and `4x-3y-25=0`
which are the equations of tangents.