Why am I getting only one equation of tangent given a point outside circle when using condition of tangency?

Question

I was given an equation to a circle:

$ x^{2}+y^{2}-6x+4y-3=0 $

and a point outside it $(7,4)$

Using slope-point form of a line, I wrote the equation of line as: $mx+y-7m-4$

Then using condition of tangency,

$\left| \dfrac{-4m-6}{\sqrt{m^{2}+1}}\right| =4$

After squaring an solving, I get: $m=-20/48$ (only one value), Please tell me what I did wrong

Answer 1

HINT

I get $$\frac{y-7}{x-4}=m~\text{ or}$$ $$ y-mx+4m-7=0\tag 1$$ You got $$y+mx-7m-4=0 \tag2$$

If you plug in equn 1) into the circle equation and letting the determinant of quadratic equn be zero for tangent condition, i.e., to eliminate x there will be two y values,

and to eliminate y there will be two x values.

Connecting these two points to $(7,4)$ you get two tangents.

Answer 2

By Thale's Theorem, a circle centered at the midpoint between the center of the circle and the target point having the center on it will intersect the original circle at points of tangency.

$x^2+y^2-6x+4y-3=0\implies (x-3)^2+(y+2)^2=16$

center $(3,-2)$

target point $(7,4)$

mid point $(5,1)$

$(x-5)^2+(y-1)^2=13$

$2(2x-8)+3(2y+1)=3$

$4x-16+6y=0$

$y=-2x/3+8/3$

$(x-3)^2+(-2x/3+14/3)^2=16$

$x^2-6x+9+4x^2/9+196/9-56x/9=16$

$13x^2/9-110x/9+133/9=0$

$x=\frac{110\pm 72}{26}=\{7,19/13\}$