Task B-4.7. A chord of length $6 \sqrt{3}$ is drawn through the point $M(6,2)$ inside the circle $(x-3)^{2}+(y-4)^{2}=36$. Write the equation of the line on which this chord lies.
## First Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_bde1d6b20cdcb90302c4g-23.jpg?height=1014&width=1152&top_left_y=393&top_left_x=541)

Sketch

(2 points)

The required lines are tangents drawn from the point \( M(6,2) \) to a concentric circle of the given circle with center at \( S(3,4) \) and radius

\[
r=\sqrt{6^{2}-27}=3
\]

If the line \( y=k x+l \) is a tangent passing through the point \( M(6,2) \), then:

\[
\begin{aligned}
& 2=6 k+l \text { i.e. } l=2-6 k \\
& (k \cdot 3-4+l)^{2}=3^{2}\left(1+k^{2}\right) \ldots \text { (tangency condition for the circle). }
\end{aligned}
\]

The solution to the obtained system is \( k=\frac{5}{12} \). Then,

\[
l=2-6 \cdot \frac{5}{12}=-\frac{1}{2}
\]

so one solution is the line \( y=\frac{5}{12} x-\frac{1}{2} \) or \( 5 x-12 y-6=0 \).

Since we can always draw two tangents from a point outside the circle, the other solution is a line parallel to the \( y \)-axis, which is the line \( x=6 \).

(2 points)