The tangents to the circle `x^2+y^2-4x+2y+k=0` at `(1, 1)` is `x-2y+1=0` then `k=`

The tangent to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

The tangent to the circle x^(2)+y^(2)-4x+2y+k=0 at (1,1) is x-2y+1=0 then k=

The equation of the tangent to the circle x^(2)+y^(2)-4x+4y-2=0 at (1,1) is

The equation of the tangent to the circle x^(2)+y^(2)-4x+4y-2=0 at (1,1) is

If y+c=0 is a tangent to the circle x^(2)+y^(2)-6x-2y+1=0 at (a, 4), then

If y+c=0 is a tangent to the circle x^2+y^2-6x-2y+1=0 at (a,4) then

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

The power of (1,1) with respect to the circle x^(2)+y^(2)-4x+3y+k=0 is 3 then k is

Find the equation of the tangent to the circle x^2+y^2-4x-6y-12=0 at (-1,-1)