Math Problem Statement

A circle with center O has a tangent PQ and a secant PR. PQ = 8 cm and PR = 12 cm. Calculate the radius of the circle.

Solution

The problem involves a tangent-secant theorem with a circle. Here is the given information extracted from the image:

  1. There is a circle with center OO and a point QQ outside the circle.
  2. PQPQ is a tangent to the circle at point PP, and PRPR is a secant line that intersects the circle at RR and QQ.
  3. The length of PQPQ (the tangent) is given as 88 cm.
  4. The length of PRPR (the secant segment from PP to RR) is given as 1212 cm.
  5. We need to find the radius rr of the circle.

Solving the Problem

The Tangent-Secant Theorem states that:

PQ2=PRPQPQ^2 = PR \cdot PQ

Given that PQ=8PQ = 8 cm and PR=12PR = 12 cm, substitute these values into the formula:

82=12PQ8^2 = 12 \cdot PQ

64=12PQ64 = 12 \cdot PQ

PQ=6412PQ = \frac{64}{12}

PQ=1635.33 cmPQ = \frac{16}{3} \approx 5.33 \text{ cm}

Now, to find the radius rr of the circle, we need to recognize that the length PRPR is composed of the sum of two parts: the external part (which is PQ=8PQ = 8 cm) and the internal part (from PP to the point of tangency QQ).

The radius can be derived by using the following relationship:

Using the formula: PQ2=PRPQ=82=(8+x)xPQ^2 = PR \cdot PQ = 8^2 = (8 + x) \cdot x Where xx is the distance of PQPQ, hence:

64=(8+x)x64 = (8 + x) \cdot x

x2+8x64=0x^2 + 8x - 64 = 0

Solving the quadratic equation:

Using the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x=8±824(1)(64)2(1)x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-64)}}{2(1)}

x=8±64+2562x = \frac{-8 \pm \sqrt{64 + 256}}{2}

x=8±3202x = \frac{-8 \pm \sqrt{320}}{2}

x=8±16202x = \frac{-8 \pm \sqrt{16 \cdot 20}}{2}

x=8±452x = \frac{-8 \pm 4\sqrt{5}}{2}

x=4±25x = -4 \pm 2\sqrt{5}

Now since xx must be positive

x=4+25x = -4 + 2\sqrt{5}

Now using the value of x the radius of the circle is:

r=163r = \frac{16}{3}

Answer Choices

The answer choice is closest to r=1635.33 cmr = \frac{16}{3} \approx 5.33 \text{ cm}, and none of the options provided matches this value.

Follow-up Questions and Tips

  1. What is the Tangent-Secant theorem, and how is it applied in geometry?
  2. Can you explain how the quadratic formula is derived?
  3. What are some other practical applications of the tangent-secant theorem?
  4. How does this problem illustrate the relationship between tangents and secants in a circle?
  5. How would the problem change if the tangent-secant theorem wasn't applicable?

Tip: Always check your units and double-check calculations, especially when using formulas involving squares or roots.

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Math Problem Analysis

Mathematical Concepts

Geometry
Circle Theorems
Tangent-Secant Theorem
Quadratic Equations

Formulas

Tangent-Secant Theorem: PQ^2 = PR * PQ
Quadratic equation formula: ax^2 + bx + c = 0

Theorems

Tangent-Secant Theorem

Suitable Grade Level

Grades 9-11