If a circle of radius $R$ passes through the origin $O$ and intersects the coordinate axes at $A$ and $B$, then the locus of the foot of perpendicular from $O$ on $AB$ is :

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To solve this problem, we need to find the equation representing the locus of the foot of the perpendicular from the origin $O$ to the line segment $AB$. The circle's center is $(h, k)$ and has a radius $R$. It can be defined by the equation:

$(x - h)^2 + (y - k)^2 = R^2$

Since the circle passes through the origin $O (0, 0)$, substituting this point into the circle's equation provides:

$h^2 + k^2 = R^2$

The circle intersects the axes at $A (a, 0)$ and $B (0, b)$. Substituting these points into the circle's equation gives:

For $(a, 0)$: $(a-h)^2 + k^2 = R^2$
For $(0, b)$: $h^2 + (b-k)^2 = R^2$

The equation of line $AB$ can be expressed using point-slope form:

$\frac{x}{a} + \frac{y}{b} = 1$

We seek the foot of the perpendicular from $O$ to $AB$. In standard coordinates, we can use the formula for the foot of a perpendicular from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$:

The coordinates of the foot are:

$\left(\frac{B(Bx_0 - Ay_0) - A C}{A^2 + B^2}, \frac{A(-Bx_0 + Ay_0) - B C}{A^2 + B^2}\right)$

Substituting $x_0 = 0, y_0 = 0, A = \frac{1}{a}, B = \frac{1}{b}, C = -1$ gives:

$\left( \frac{-a}{b^2 + a^2}, \frac{-b}{b^2 + a^2} \right)$

This is $(x, y)$. Hence, the locus:

$x^2 + y^2 = \left(\frac{1}{a^2} + \frac{1}{b^2} \right)$

With $a^2 + b^2 = R^2$, equating gives the equation in $(x, y)$:

$(x^2 + y^2)^3 = 4R^2x^2y^2$

This matches the correct answer.

Answer 2