Answer:
![r= \frac { \sqrt{ {l}^(2) + {m}^(2) } }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vlabppyk53zznokn3gmeib59fo6cxbv94k.png)
Explanation:
The given circle has equation
![{x}^(2) + {y}^(2) + 2gx + 2fy + c = 0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/kiq3y3ic5x6kh8kjgan42unnh7t7cj491u.png)
This circle has centre (-g,-f).
For the line
![lx + my + n = 0](https://img.qammunity.org/2021/formulas/mathematics/middle-school/udmzzlaf3ens37cqzzvda73uga3xb40tdp.png)
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:
![d = \frac { \sqrt{ {a}^(2) + {b}^(2) } }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2dlm8hkdzfp9g0u0pqv415rto3zvmg9zuj.png)
We substitute the center and radius to get:
![r=\frac { \sqrt{ {l}^(2) + {m}^(2) } }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/ruu8v4eydpqw93559vjusare0d60x33300.png)
We simplify to get;
![r= \frac { \sqrt{ {l}^(2) + {m}^(2) } }](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vlabppyk53zznokn3gmeib59fo6cxbv94k.png)