What is the equation of the tangent to the circle $x^{2}+y^{2}=4$ that makes an angle of $\alpha=143^{\circ} 7^{\prime} 48.2^{\prime \prime}$ with the $X$-axis?

What is the equation of the tangent that passes through the point $P\left(0,-\frac{10}{3}\right)$?
$1^{\circ} \mathrm{Az}$ tangent equation

$$
\pm x \cos \beta \pm y \sin \beta=r
$$

where

$$
\beta=\alpha-90^{\circ}=53^{\circ} 7^{\prime} 48.2^{\prime \prime}
$$

But $\cos \beta=0.6$ and $\sin \beta=0.8 \mathrm{~s}$, so the sought equations are:

$$
3 x+4 y=10 \quad \text { and } \quad 3 x+4 y=-10
$$

(Pál Hajdu, Budapest.)

$2^{\circ} \mathrm{Az}$ tangent equation:

$$
x x_{1}+y y_{1}=4
$$

Substituting the given values 0 and $-\frac{10}{3}$ for $x$ and $y$:

$$
y_{1}=-\frac{6}{5}
$$

Substituting this value into the circle equation, we get:

$$
x_{1}= \pm \frac{8}{5}
$$

Therefore, the equations of the tangents that can be drawn from point $P$ are

$$
4 x-3 y=10 \quad \text { and } \quad 4 x+3 y=-10
$$