Question

Let \(C_1\) and \(C_2\) be concentric circles such that the diameter of \(C_1\) is \(2\) cm longer than that of \(C_2\). If a chord of \(C_1\) has length \(6\) cm and is a tangent to \(C_2\), then the diameter, in cm, of \(C_1\) is

Answer 1

Given: \[ d + 2 = D \quad \text{and} \quad r + 1 = R \] In the provided diagram:

• \(OT = r\) (radius of the inner circle)
• \(OB = r + 1\) (radius of the outer circle)

Step 1: Analyze the Geometric Setup

Since \(AB\) is a common tangent to both circles and \(OT \perp AB\), a right triangle \(OTB\) is formed. The base \(TB = \frac{6}{2} = 3\) because the tangent segment \(AB = 6\) cm is bisected at \(T\).

Step 2: Apply the Pythagorean Theorem to \(\triangle OTB\)

In the right triangle: \[ OT^2 + TB^2 = OB^2 \] Substituting the given values: \[ r^2 + 3^2 = (r + 1)^2 \] Expanding and simplifying: \[ r^2 + 9 = r^2 + 2r + 1 \Rightarrow 9 = 2r + 1 \Rightarrow 2r = 8 \Rightarrow r = 4 \]

Step 3: Calculate the Diameter of the Larger Circle

The radius of the larger circle is \(R = r + 1 = 4 + 1 = 5\). The diameter \(D\) is twice the radius: \(D = 2R = 2 \times 5 = \boxed{10\text{ cm}}\)

Final Answer:

The diameter of the larger circle is: \[ \boxed{10 \text{ cm}} \]