To determine the point of intersection where the tangent line $L$ touches the circle, we start with the given information:

1. The circle is centered at the origin $(0,0)$ with a radius of 10.
2. The tangent line $L$ is parallel to the line $y = 5x + 7$ and has a negative y-intercept.

Since the tangent line is parallel to $y = 5x + 7$, it has the same slope. Therefore, the slope of the tangent line $L$ is 5. We can express the equation of $L$ in the form: $y = 5x + c$

Given that the y-intercept $c$ is negative, we need to find the value of $c$ such that the line is tangent to the circle.

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must equal the radius of the circle. The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

For our line $y = 5x + c$, we rewrite it in the standard form: $5x - y + c = 0$

Here, $A = 5$, $B = -1$, and $C = c$. The center of the circle is at $(0, 0)$, so $x_1 = 0$ and $y_1 = 0$. The distance from the center to the line is: $d = \frac{|5(0) - 1(0) + c|}{\sqrt{5^2 + (-1)^2}} = \frac{|c|}{\sqrt{25 + 1}} = \frac{|c|}{\sqrt{26}}$

Since the line is tangent to the circle, this distance $d$ must equal the radius of the circle, which is 10. Therefore: $\frac{|c|}{\sqrt{26}} = 10$ $|c| = 10 \sqrt{26}$

Since $c$ is negative: $c = -10 \sqrt{26}$

Thus, the equation of the tangent line $L$ is: $y = 5x - 10 \sqrt{26}$

To find the point of intersection, we need to find the point $(x_0, y_0)$ where the tangent touches the circle. This point satisfies both the equation of the circle and the tangent line. Substituting $y = 5x - 10 \sqrt{26}$ into the circle's equation $x^2 + y^2 = 100$:

$x^2 + (5x - 10\sqrt{26})^2 = 100$ $x^2 + 25x^2 - 100x\sqrt{26} + 100(26) = 100$ $26x^2 - 100x\sqrt{26} + 2600 = 100$ $26x^2 - 100x\sqrt{26} + 2500 = 0$

Solving this quadratic equation for $x$: $x = \frac{100\sqrt{26} \pm \sqrt{(100\sqrt{26})^2 - 4 \cdot 26 \cdot 2500}}{2 \cdot 26}$ $x = \frac{100\sqrt{26} \pm \sqrt{10000 \cdot 26 - 260000}}{52}$ $x = \frac{100\sqrt{26} \pm \sqrt{260000 - 260000}}{52}$ $x = \frac{100\sqrt{26}}{52}$ $x = \frac{50\sqrt{26}}{26}$ $x = \frac{25\sqrt{26}}{13}$

Now solving for $y$: $y = 5 \left( \frac{25\sqrt{26}}{13} \right) - 10\sqrt{26}$ $y = \frac{125\sqrt{26}}{13} - 10\sqrt{26}$ $y = \frac{125\sqrt{26} - 130\sqrt{26}}{13}$ $y = \frac{-5\sqrt{26}}{13}$

Thus, the point of intersection where the tangent line touches the circle is: $\left( \frac{25\sqrt{26}}{13}, \frac{-5\sqrt{26}}{13} \right)$

Would you like more details or have any questions about this solution? Here are some related questions for further exploration:

1. How would the equation change if the circle's center was at a different point, say $(a, b)$?
2. What would be the result if the slope of the tangent line were different?
3. How can we derive the perpendicular distance formula from a point to a line?
4. What if the y-intercept of the tangent line was positive?
5. Can we use another method to find the point of tangency without solving a quadratic equation?

Tip: Understanding the geometric properties of circles and lines can simplify solving tangency problems, often reducing the need for complex algebraic manipulations.