How to find the equation of tangents from a given point to general circle

Question

I know the formula for equation of tangents for a standard circle $x^2+y^2=a^2$ which is $$y=mx\pm a\sqrt{1+m^2}$$ From where we can find the slope of the tangents. Is there any such equation for a circle of the form $x^2+y^2+2gx+2fy+c=0$

Answer 1

For all real values (including infinite) of $m,$ the straight line will be tangent to the circle.

Hint:

For any conic in 2D $$ax^2+2hxy+by^2+2gx+2fy+c=0$$,

let $y=mx+d$ be the tangent.

Replace the value of $y$ in terms of $x$ to form a quadratic equation in $x$.

Each root will represent the abscissa of intersection.

For tangency, the two abscissas must coincide, i.e., the roots must be same

Answer 2

In the standard equation of tangent to circle $y=mx\pm a\sqrt{1+m^2}$. Replace $(x,y)$ with $(x+g,y+f)$ and radius $a$ with $r=\sqrt{g^2+f^2-c}$. $$. $$ Hence equation of tangent to circle $x^2+y^2+2gx+2fy+c=0$ is $$y+f=m(x+g) \pm r\sqrt{1+m^2}$$