Answer:
[tex]r= \frac{ |c - lg - fm| }{ \sqrt{ {l}^{2} + {m}^{2} } } [/tex]
Step-by-step explanation:
The given circle has equation
[tex] {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0[/tex]
This circle has centre (-g,-f).
For the line
[tex]lx + my + n = 0[/tex]
to be a tangent to this circle, the perpendicular distance from the center to this line must be equal to the radius of the circle.
This is given by:
[tex]d = \frac{ |ax + by + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } [/tex]
We substitute the center and radius to get:
[tex]r=\frac{ |l \times - g + m \times - f+ c| }{ \sqrt{ {l}^{2} + {m}^{2} } } [/tex]
We simplify to get;
[tex]r= \frac{ |c - lg - fm| }{ \sqrt{ {l}^{2} + {m}^{2} } } [/tex]
