If $P(-9,-1)$ is a point on the circle $x^2+y^2+4x+8y-38=0$, then find equation of the tangent drawn at the other end of the diameter drawn through $P$.

We have a circle, $x^2+y^2+4x+8y-38=0 \Rightarrow {\left(x+2\right)}^2+{\left(y+4\right)}^2=58$

with centre$\left(-2, -4\right)$ and radius $\sqrt{58}$

Now, there is a point $P\left(-9, -1\right)$ on the given circle.

Let there is a point $Q\left(x_1, y_1\right)$ on the other end of the diameter drawn through point $P.$
Hence, $\frac{x_1-9}{2}=-2 \Rightarrow x_1=5$ and $\frac{y_1-1}{2}=-4\Rightarrow y_1=-7$

So, the other end of the diameter through $P$ is $Q\left(5, -7\right)$.

Now, the slope of the normal to the point $Q$ is the slope of the diameter $PQ$.

i.e. Slope of $PQ=\frac{-1-\left(-7\right)}{-9-5}=-\frac{6}{14}=-\frac{3}{7}$

So, the slope of the tangent through point $P$ is $=-\frac{1}{-{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}=\frac{7}{3}$

Hence, the equation of the tangent through point $Q$ is .

$y+7=\frac{7}{3}\left(x-5\right) \Rightarrow 7x-3y=56$