A circle C0 with radius 1 unit touches both the axes and as well as line L1 through P(0, 4) and (x1, 0). Now a circle C1 of radius 1 unit is drawn touching the x axis, line L1 and another line L2 passing through P. L2 intersects x axis at (x2, 0) and this process is repeated n times as shown Then which of the following is/are true?

For Cn circle let center be (an,1)and it touches the line Ln and Ln+1i.e xxn+y4=1,xxn+1+y4=1⇒4αn+xn−4xn16+xn2=1⇒4αn+xn+1−4xn+116+xn+12=−1…… (i) 3xn+1−xn=16+xn2+16+xn+12 Let xn=4cot⁡θn0<θn<π2⇒tan⁡θn+12=12tan⁡θn2 As x1=3⇒tan⁡θ12−12⇒tan⁡θn2=12n⇒xn=22n−12n−1x1=152Putting is (i), we get α2=314limn→∞ xn2n=limn→∞ 22n−12n−12n=2

A circle C0 with radius 1 unit touches both the axes and as well as line L1 through P(0, 4) and (x1, 0). Now a circle C1 of radius 1 unit is drawn touching the x axis, line L1 and another line L2 passing through P. L2 intersects x axis at (x2, 0) and this process is repeated n times as shown Then which of the following is/are true?

For Cn circle let center be (an,1)and it touches the line Ln and Ln+1i.e xxn+y4=1,xxn+1+y4=1⇒4αn+xn−4xn16+xn2=1⇒4αn+xn+1−4xn+116+xn+12=−1…… (i) 3xn+1−xn=16+xn2+16+xn+12 Let xn=4cot⁡θn0<θn<π2⇒tan⁡θn+12=12tan⁡θn2 As x1=3⇒tan⁡θ12−12⇒tan⁡θn2=12n⇒xn=22n−12n−1x1=152Putting is (i), we get α2=314limn→∞ xn2n=limn→∞ 22n−12n−12n=2

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A circle C₀ with radius 1 unit touches both the axes and as well as line L₁ through P(0, 4) and (x₁, 0). Now a circle C₁ of radius 1 unit is drawn touching the x axis, line L₁ and another line L₂passing through P. L₂ intersects x axis at
(x₂, 0) and this process is repeated n times as shown Then which of the following is/are true?

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$x_{2} = \frac{15}{2}$

The centre of the circle C₂ is $(\frac{15}{2}, 1)$

The centre of the circle C₂is $(\frac{31}{4}, 1)$

x₁ = 3

answer is B, C, D.

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Detailed Solution

For C_n circle let center be (a_n,1)and it touches the line L_n and L_n+1
i.e $\frac{x}{x_{n}} + \frac{y}{4} = 1, \frac{x}{x_{n + 1}} + \frac{y}{4} = 1$
$\begin{array}{l} \Rightarrow \frac{4 α_{n} + x_{n} - 4 x_{n}}{\sqrt{16 + x_{n}^{2}}} = 1 \\ \Rightarrow \frac{4 α_{n} + x_{n + 1} - 4 x_{n + 1}}{\sqrt{16 + x_{n + 1}^{2}}} = - 1 \dots \dots (i) \\ 3 (x_{n + 1} - x_{n}) = \sqrt{16 + x_{n}^{2}} + \sqrt{16 + x_{n + 1}^{2}} \\ Let x_{n} = 4 \cot θ_{n} 0 < θ_{n} < \frac{π}{2} \end{array}$
$\begin{array}{l} \Rightarrow \tan (\frac{θ_{n + 1}}{2}) = \frac{1}{2} \tan (\frac{θ_{n}}{2}) \\ As x_{1} = 3 \Rightarrow \tan \frac{θ_{1}}{2} - \frac{1}{2} \Rightarrow \tan (\frac{θ_{n}}{2}) = \frac{1}{2^{n}} \\ \Rightarrow x_{n} = (\frac{2^{2 n} - 1}{2^{n - 1}}) x_{1} = \frac{15}{2} \end{array}$
Putting is (i), we get $α_{2} = \frac{31}{4}$
$lim_{n \to \infty} \frac{x_{n}}{2^{n}} = lim_{n \to \infty} \frac{2^{2 n} - 1}{2^{n - 1} 2^{n}} = 2$