One end of the diameter of the circle $x^2+y^2-6 x-5 y-1=0$ is $(-1,3)$, then the equation of the tangent at…
One end of the diameter of the circle $x^2+y^2-6 x-5 y-1=0$ is $(-1,3)$, then the equation of the tangent at the other end of the diameter is
$8 x+y-58=0$
$8 x-2 y-52=0$
$8 x-y-54=0$
$8 x+2 y-60=0$
Solution
If $\left(x_1, y_1\right)$ is one end of a diameter of the circle $x^2+y^2+2 \mathrm{~g} x+2 \mathrm{f} y+\mathrm{c}=0$, then the other end is $\left[-\left(x_1+2 \mathrm{~g}\right),-\left(y_1+2 \mathrm{f}\right)\right]$
$\therefore \quad$ The other end of $x^2+y^2-6 x-5 y-1=0$ is $[-(-1-6),-(3-5)]$ i.e. $(7,2)$
$\therefore \quad$ Equation of tangent at $(7,2)$ is
$\begin{aligned}
& 7 x+2 y-3(x+7)-\frac{5}{2}(y+2)-1=0 \\
& \Rightarrow 8 x-y-54=0
\end{aligned}$