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, \(d+2=D\)
\(⇒ r+1 = R\)
In the figure \(OT = r\) and \(OB = r+1\)
\(OT ⊥ AB\) as \(AB\) is the tangent
OT bisects AB i.e., \(TB=\frac 62 = 3\)
Now, in ΔOTB, \(OT^2+TB^2 = OB^2\)
∴ \(r^2+32=(r+1)^2\)
\(⇒ r=4\)
∴ \(D=2(R)=2(r+1)=10\) cm

So, the correct option is (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.