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

We are given that: \[ d + 2 = D \quad \text{and} \quad r + 1 = R \] In the given figure:

• \(OT = r\) (radius of smaller circle)
• \(OB = r + 1\) (radius of larger circle, since it's 1 unit more)

 

Step 1: Understand the Geometry

Since \(AB\) is a common tangent to both circles, and \(OT\) is perpendicular to \(AB\), it forms a right triangle with base \(TB = \frac{6}{2} = 3\), because \(AB = 6\) cm and is bisected at point \(T\).

Step 2: Apply Pythagoras Theorem in Triangle \( \triangle OTB \)

Using the right triangle: \[ OT^2 + TB^2 = OB^2 \] Substitute the known values: \[ r^2 + 3^2 = (r + 1)^2 \] Simplifying: \[ r^2 + 9 = r^2 + 2r + 1 \Rightarrow 9 = 2r + 1 \Rightarrow 2r = 8 \Rightarrow r = 4 \]

Step 3: Find the Required Diameter

\[ R = r + 1 = 4 + 1 = 5 \Rightarrow D = 2R = 2 \times 5 = \boxed{10\text{ cm}} \]

Final Answer:

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

Answer 2

Let the radius of the smaller circle \( R_1 \) be \( r \), and the radius of the larger circle \( R_2 \) be \( r + 1 \).

Using the Pythagoras Theorem

From the given diagram, the distance from the center of the smaller circle to the point of tangency is \( r \), and from the center of the larger circle is \( r + 1 \). The half-length of the tangent between the two contact points is 3 cm.

Applying the Pythagoras theorem: \[ (r + 1)^2 = r^2 + 3^2 \] Simplifying: \[ r^2 + 2r + 1 = r^2 + 9 \Rightarrow 2r = 8 \Rightarrow r = 4 \]

Calculating the Diameter

The radius of the larger circle is: \[ R = r + 1 = 4 + 1 = 5 \] So, the diameter of the larger circle \( C_1 \) is: \[ D = 2R = 2 \times 5 = \boxed{10\ \text{cm}} \]

Final Answer:

10 cm