Question

Let the locus of the centre (α, β), β> 0, of the circle which touches the circle x² +(y – 1)² = 1 externally and also touches the x-axis be L. Then the area bounded by L and the line y = 4 is :

• A. \(\frac{\sqrt{32}}{3}\)
• B. \(\frac{40\sqrt2}{3}\)
• C. \(\frac{64}{3}\)
• D. \(\frac{32}{3}\)

Answer 1

The Correct Option is C

To find the area bounded by the locus L and the line \(y = 4\), we first determine the nature and equation of locus L, which represents the path of the center of all possible circles that touch both the circle \(x^2 + (y - 1)^2 = 1\) and the x-axis.

The given circle \(x^2 + (y - 1)^2 = 1\) has its center at \((0, 1)\) and a radius of \(1\).

Let the center of the required circle be \((\alpha, \beta)\), and its radius be \(r\). Since it touches the x-axis, we have \(\beta = r\).

Additionally, since this circle touches the given circle externally, the distance between their centers equals the sum of their radii:

r + 1 = \sqrt{(0 - \alpha)^2 + (1 - \beta)^2}

Simplifying, we have:

r + 1 = \sqrt{\alpha^2 + (\beta - 1)^2}

We know \(r = \beta\), so substituting:

\beta + 1 = \sqrt{\alpha^2 + (\beta - 1)^2}

Squaring both sides:

(\beta + 1)^2 = \alpha^2 + (\beta - 1)^2

Expanding and simplifying gives:

\beta^2 + 2\beta + 1 = \alpha^2 + \beta^2 - 2\beta + 1

Thus, \(4\beta = \alpha^2\).

This means the locus of points \((\alpha, \beta)\) is a parabola: \(\beta = \frac{\alpha^2}{4}\)

To find the area bounded by this parabola and the line \(y = 4\), we find the x-values where the parabola intersects \(y = 4\):

\frac{\alpha^2}{4} = 4 \Rightarrow \alpha^2 = 16 \Rightarrow \alpha = \pm 4

The area under one arc of the parabola from \(\alpha = -4\) to \(\alpha = 4\) is given by:

\int_{-4}^{4} \left(4 - \frac{\alpha^2}{4}\right) d\alpha

Calculating, we get:

= \int_{-4}^{4} 4\,d\alpha - \frac{1}{4}\int_{-4}^{4} \alpha^2\,d\alpha

= [4\alpha]_{-4}^{4} - \frac{1}{4}\left[\frac{\alpha^3}{3}\right]_{-4}^{4}

= (16 - (-16)) - \frac{1}{4}\left(\frac{64}{3} - \left(-\frac{64}{3}\right)\right)

= 32 - \frac{1}{4}\left(\frac{128}{3}\right)

= 32 - \frac{32}{3}

= \frac{96}{3} - \frac{32}{3}

= \frac{64}{3}

Therefore, the area bounded by the locus L and the line \(y = 4\) is \( \frac{64}{3} \).