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, \(d+2=D ⇒ r+1 = R\)
In the figure OT = \(r\) and OB = \(r+1\)
OT ⊥ AB as AB is the tangent
OT bisects AB i.e., TB = \(\frac{6}{2} = 3\)
Now, in \(ΔOTB, OT^2+TB^2 = OB^2\)
\(∴ r^2+3^2=(r+1)^2 ⇒ r=4\)
\(∴ D=2(R)=2(r+1)=10cm\)
So, the correct answer is 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.