Problem 75 - Algebra and Trigonometry with Analytic Geometry

Question

From an exterior point \(P\) that is \(h\) units from a circle of radius \(r\), a tangent line is drawn to the circle (see the figure). Let \(y\) denote the distance from the point \(P\) to the point of tangency \(T\). (a) Express \(y\) as a function of \(h\). (Hint: If \(C\) is the center of the circle, then \(P T\) is perpendicular to \(C T\).) (b) If \(r\) is the radius of Earth and \(h\) is the altitude of a space shuttle, then \(y\) is the maximum distance to Earth that an astronaut can see from the shuttle. In particular, if \(h=200 \mathrm{mi}\) and \(r=4000 \mathrm{mi}\), approximate \(y\).

Step-by-Step Solution

Answer

\( y = \sqrt{2rh + h^2} \) and for \( h = 200 \) mi, \( y \approx 1280.62 \) mi.

Step 1: Understand the Geometric Setup

Consider a circle with center \( C \) and radius \( r \). The point \( P \) is an external point located \( h \) units away from \( C \). The line connecting \( P \) to \( C \) is of length \( r + h \). There is a tangent line drawn from \( P \) to the circle, touching at point \( T \), and we need to find the length \( y = PT \).

Step 2: Recognize the Right Triangle

Identify the right triangle \( \triangle PCT \), where \( PC = r + h \) is the hypotenuse, \( CT = r \) is one leg, and \( PT = y \) is the other leg, the tangent we want to find. Since \( PT \perp CT \), \( \angle PTC = 90^\circ \).

Step 3: Apply the Pythagorean Theorem

Using the Pythagorean theorem for \( \triangle PCT \), we have: \[ (PT)^2 + (CT)^2 = (PC)^2 \]Substitute the known lengths: \[ y^2 + r^2 = (r + h)^2 \]

Step 4: Solve for \( y \) in terms of \( h \)

Rearrange the equation from Step 3 to solve for \( y \):\[ y^2 = (r + h)^2 - r^2 \]Simplify the expression:\[ y^2 = (r^2 + 2rh + h^2) - r^2 = 2rh + h^2 \]So, \( y = \sqrt{2rh + h^2} \).

Step 5: Calculate \( y \) for the Given Values

For \( r = 4000 \) mi and \( h = 200 \) mi, substitute into the equation found in Step 4:\[ y = \sqrt{2 \times 4000 \times 200 + 200^2} \]\[ y = \sqrt{1600000 + 40000} = \sqrt{1640000} \]Calculate the approximate value of \( y \): \( y \approx 1280.62 \) mi.

Key Concepts

Geometric Problem-Solving

In this scenario, we are dealing with the problem of finding the length of a tangent line drawn from an external point to a circle. This type of problem is common in geometry, especially when learning about circles and their properties. The fundamental idea here is to understand the spatial arrangement of the various elements involved:

• The circle with a given radius,
• The external point from which the tangent is drawn,
• The center of the circle,
• And the tangent line itself.

To solve the problem, one must be familiar with different geometric properties. Key pointers include understanding how an external point, the circle's center, and the tangent intermingle to form a right triangle. This setup is pivotal, illustrating the importance of geometric intuition and spatial reasoning when approaching such exercises. Having a clear mental image or a diagram strengthens one's ability to solve the problem accurately.

Pythagorean Theorem

The Pythagorean Theorem is a vital tool in this exercise. It applies to right triangles and relates the lengths of the sides: the hypotenuse and the two other legs. In this problem, the diagonal line from the point to the center of the circle forms the hypotenuse, while the radius of the circle and our tangent line form the two other legs.The theorem states: \[ a^2 + b^2 = c^2 \]where:

• \( a \) and \( b \) are the legs of the triangle,
• \( c \) is the hypotenuse.

In our setup, as \( PC = r + h \) is the hypotenuse, and \( r \) and \( y \) are the legs, it transforms into:\[ y^2 + r^2 = (r + h)^2 \]This equation helps us derive the needed tangent length \( y \) by rearranging it and solving for \( y \). The simplification of this equation is straightforward once you substitute the known values and solve step by step, showing the powerful application of a fundamental theorem in practical, applied settings.

Right Triangle

A right triangle plays a crucial role in this geometric exercise. Specifically, it helps us apply mathematical concepts like the Pythagorean Theorem. The configuration forms a right triangle, labeled as \( \triangle PCT \), where the hypotenuse \( PC \), leg \( CT \), and the tangent line \( PT \) create a 90-degree angle at \( T \).This right triangle is established because the tangent line is always perpendicular to the radius at the point of tangency on the circle. This rule is consistent for all circles and tangents and serves as the basis for many geometrical problem-solving tasks.Recognizing a right triangle allows us to effectively calculate unknown lengths and verify our solution's accuracy through known relationships. Understanding this principle can greatly simplify the calculation process. The right triangle framework is not just a geometric shape but a key provider of the information necessary to solve complex geometric problems.