Find the equation of the tangent to the circle $x^2 + y^2 + 8x + 10y - 89 = 0$ at the point $\left(-1, 6\right)$.
Equation of given circle: $\;$ $x^2 + y^2 + 8x + 10y - 89 = 0$
Comparing with the standard equation of circle $\;$ $x^2 + y^2 + 2gx + 2fy + c = 0$ $\;$ gives
$g = 4$, $\;$ $f = 5$, $\;$ $c = -89$
Given point: $\;$ $\left(x_1, y_1\right) = \left(-1, 6\right)$
Equation of tangent to the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ at any given point $\left(x_1, y_1\right)$ is
$xx_1 + yy_1 + g \left(x + x_1\right) + f \left(y + y_1\right) + c = 0$
$\therefore \;$ Required equation of tangent is
$-x + 6y + 4 \left(x - 1\right) + 5 \left(y + 6\right) - 89 = 0$
i.e. $\;$ $-x + 6y + 4x -4 + 5y + 30 - 89 = 0$
i.e. $\;$ $3x + 11y - 63 = 0$