If the equation of tangent to the circle `x^2+y^2-2x+6y-6=0` and parallel to `3x-4y+7=0` is `3x-4y+k=0` then the values of k are

Options

A. 5,-35
B. `-5,35`
C. `7,-32`
D. `-7,32`

Answer

A

Explanation

We have, equation of cricle is `x^(2)+y^(2)-2x+6y-6=0.`
`rArrg=-1,f=3,c=-6`
`therefore radiuds (r)=sqrt(1+9+6=4)`
and centre=(1,3) now, distance of tangent 3x-4y+k=0, from the centre is `4=(3+12+k)/(sqrt(25))`
`rArr 4=(k+15)/(+-5)`
`rArr +-20=k+15`
`therefore k=5,-35`