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
[This Question was asked as TITA]

• A. 10 cm
• B. 12 cm
• C. 15 cm
• D. 18 cm

Answer 1

The Correct Option is A

Given: \( d + 2 = D \Rightarrow r + 1 = R \)

In the figure, the small circle has radius \( r \), so: 

• \( OT = r \) (radius of the smaller circle)
• \( OB = r + 1 = R \) (radius of the larger circle)
• Since AB is a tangent to the smaller circle, it is perpendicular to radius OT at point T.
• OT ⊥ AB
• Also, OT bisects AB, hence: \( TB = \frac{6}{2} = 3 \, \text{cm} \)

Now, using the right triangle \( \triangle OTB \):

\[ OT^2 + TB^2 = OB^2 \]

Substitute values:

\[ r^2 + 3^2 = (r + 1)^2 \]

\[ r^2 + 9 = r^2 + 2r + 1 \]

Cancel \( r^2 \) on both sides:

\[ 9 = 2r + 1 \Rightarrow 2r = 8 \Rightarrow r = 4 \]

So, \( R = r + 1 = 5 \)

Then, diameter of the larger circle is: \[ D = 2R = 2 \times 5 = \mathbf{10 \, \text{cm}} \]

Correct option: (A) 10 cm

Answer 2

Let the radius of R₁ = r and R₂ = r + 1
Now , According to the above diagram,
By using the Pythagoras theorem, we get :
⇒ (r+1)² = (r)² + (3)²
⇒  r2 + 2r + 1 = r² + 9
⇒ 2r = 10
r = 5
Now , the diameter of C₁ is 2R
2R = 10
Therefore, the correct answer is 10 cm.