To solve the problem, we can apply the secant-tangent theorem (also known as the power of a point theorem). This theorem states that if a tangent and a secant are drawn from an external point, then the square of the length of the tangent segment is equal to the product of the entire secant length and the external secant segment.

In this case:

• $JK$ is the tangent, and its length is 27.
• $MLK$ is the secant, with $MK = ML + LK$, and $LK = 20$. We are tasked with finding $ML$.

According to the theorem:

$JK^2 = ML \times MK$

Substitute the values:

$27^2 = ML \times (ML + 20)$

This simplifies to:

$729 = ML \times (ML + 20)$

Now, we can solve this quadratic equation:

$729 = ML^2 + 20ML$ $ML^2 + 20ML - 729 = 0$

We can solve this quadratic equation using the quadratic formula:

$ML = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

• $a = 1$,
• $b = 20$,
• $c = -729$.

Substituting into the formula:

$ML = \frac{-20 \pm \sqrt{20^2 - 4(1)(-729)}}{2(1)}$ $ML = \frac{-20 \pm \sqrt{400 + 2916}}{2}$ $ML = \frac{-20 \pm \sqrt{3316}}{2}$ $ML = \frac{-20 \pm 57.6}{2}$

Taking the positive root:

$ML = \frac{-20 + 57.6}{2} = \frac{37.6}{2} = 18.8$

Thus, the length of $ML$ is approximately 18.8 units.

Would you like more details on the solution or clarification on any part? Here are five follow-up questions related to this concept:

1. What other types of circle theorems can be used with secants and tangents?
2. How would the result change if the length of the tangent increased?
3. Could we solve this geometrically instead of algebraically? How?
4. What if the secant only intersected the circle at one point?
5. Can the secant-tangent theorem be used with ellipses or other conic sections?

Tip: Always double-check the use of the quadratic formula, especially the discriminant, to avoid errors when solving real-world geometric problems.