[ Coordinate method on the plane ]

[ Properties and characteristics of the tangent ]

Form the equation of the line passing through the point $A(0 ; 7)$ and tangent to the circle $(x-15)^{2}+(y-2)^{2}=25$.

#
Let $k$ be the slope of the desired tangent line. Then the equation of the tangent line is $y-7=k(x-0)$, or $y=k x+7$. This problem reduces to finding all such numbers $k$ for which the system of equations

$$
\left\{\begin{array}{l}
(x-15)^{2}+(y-2)^{2}=25 \\
y=k x+7
\end{array}\right.
$$

has exactly one solution.

Substituting $y=k x+7$ into the first equation, after obvious simplifications, we obtain the quadratic equation

$$
\left(k^{2}+1\right) x^{2}+10 x(k-3)+225=0
$$

This equation has a unique solution if and only if its discriminant is 0, i.e.,

$$
D=10^{2}(k-3)^{2}-4 \cdot 225\left(k^{2}+1\right)=100\left(-8 k^{2}-6 k\right)=0
$$

From this, we find that $k=0$ or $k=-\frac{3}{4}$.

Therefore, the desired equation is $y=7$ or $y=-\frac{3}{4} x+7$.

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-50.jpg?height=589&width=918&top_left_y=1606&top_left_x=570)

Answer

$y=7$ or $y=-\frac{3}{4} x+7$.