Find the equation of the tangent to the circle $\;$ $x^2 + y^2 - 4x + 8y -5 = 0$ $\;$ at $\;$ $\left(2,1\right)$.
Equation of given circle: $\;\;\;$ $x^2 + y^2 - 4x + 8y - 5 = 0$
Comparing with the standard equation of circle: $\;\;\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$2g = -4$ $\implies$ $g = -2$; $\;\;\;$ $2f = 8$ $\implies$ $f = 4$; $\;\;\;$ $c = -5$
Equation of tangent to the circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ at the point $P \left(x_1, y_1\right)$ is
$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$
Given: External point $P \left(x_1, y_1\right) = \left(2,1\right)$
$\therefore$ $\;$ Equation of required tangent is
$2x + y - 2 \left(x + 2\right)+ 4 \left(y + 1\right) - 5 = 0$
i.e. $2x + y - 2x - 4 + 4y + 4 - 5 = 0$
i.e. $5y - 5 = 0$
i.e. $y - 1 = 0$