Q.The line $\dfrac{x}{a}+\dfrac{y}{b}=1$ moves in such a way that $\dfrac{1}{a^2}+\dfrac{1}{b^2}=\dfrac{1}{c^2}$, where $c$ is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^2+y^2=c^2$.

The key idea is finding the locus of a point using its geometric properties, specifically the perpendicular distance from the origin to a line.

1. Let $(h,k)$ be the foot of the perpendicular from the origin $(0,0)$ to the line $\dfrac{x}{a}+\dfrac{y}{b}=1$. The distance from the origin to $(h,k)$ is the perpendicular distance from the origin to the line.
2. Rewrite the line equation as $bx+ay-ab=0$. The perpendicular distance $p$ from the origin $(0,0)$ to this line is given by the formula:

3. Since $(h,k)$ is the foot of the perpendicular, the square of its distance from the origin is $h^2+k^2 = p^2$. $h^2+k^2 = \left(\frac{ab}{\sqrt{a^2+b^2}}\right)^2 = \frac{a^2b^2}{a^2+b^2}$ …

The problem asks us to determine the locus of a specific point: the foot of the perpendicular from the origin to a line. This line is not fixed; its intercepts $a$ and $b$ vary, but they always satisfy a given condition. Finding a locus means finding an equation that describes all possible positions of this point, independent of the varying parameters ($a$ and $b$ in this case).

Let's denote the foot of the perpendicular from the origin $O(0,0)$ to the line as $P(h,k)$. The core idea is to establish relationships between $h, k, a,$ and $b$ using the geometric properties of $P$. There are two key properties:

1. The point $P(h,k)$ lies on the given line.
2. The line segment $OP$ is perpendicular to the given line.

By leveraging these two conditions, we can derive expressions for $a$ and $b$ in terms of $h$ and $k$. Substituting these expressions into the given constraint involving $a$ and $b$ will eliminate $a$ and $b$, leaving us with an equation solely in $h$ and $k$. This equation will be the locus.

1. Identify the given line and the point: The equation of the line is given in the intercept form:

Let $P(h,k)$ be the foot of the perpendicular from the origin $O(0,0)$ to this line. We need to find the locus of $P(h,k)$.

2. Apply the condition that $P(h,k)$ lies on the line:

Since $P(h,k)$ is a point on the line, its coordinates must satisfy the line's equation. Substituting $x=h$ and $y=k$:

3. Apply the perpendicularity condition: First, find the slope of the line segment $OP$. The origin is $O(0,0)$ and $P$ is $(h,k)$.

Next, find the slope of the given line. We can rewrite $\frac{x}{a} + \frac{y}{b} = 1$ as $bx + ay = ab$, or $bx + ay - ab = 0$.
The slope of the line, $m_L$, is given by $-\frac{\text{coefficient of } x}{\text{coefficient of } y}$:

Since $OP$ is perpendicular to the line, the product of their slopes must be $-1$:

Multiplying both sides by $-1$:

4. Express $a$ and $b$ in terms of $h$ and $k$:

   We now have two equations involving $h, k, a,$ and $b$:

   (1) $\frac{h}{a} + \frac{k}{b} = 1$

   (2) $kb = ha$

   From Equation (2), we can express $a$ in terms of $b, h, k$:

Substitute this expression for $a$ into Equation (1):

To eliminate the denominators, multiply the entire equation by $kb$:

From this, we can express $b$ in terms of $h$ and $k$:

Now, substitute this expression for $b$ back into $a = \frac{kb}{h}$:

So, we have found $a$ and $b$ in terms of $h$ and $k$:

5. Use the given constraint to eliminate $a$ and $b$: The problem states that the line moves such that its intercepts satisfy the condition:

Substitute the expressions for $a$ and $b$ we just derived into this constraint:

Since the terms on the left side have a common denominator, we can combine them:

Assuming $h^2+k^2 \neq 0$ (which must be true if $c$ is a non-zero constant, as the foot of the perpendicular cannot be the origin if the line is not passing through the origin), we can cancel one factor of $(h^2+k^2)$ from the numerator and denominator:

Cross-multiplying gives:

6. State the locus: The equation $h^2+k^2 = c^2$ is the relationship satisfied by the coordinates $(h,k)$ of the foot of the perpendicular. To express the locus, we replace $h$ with $x$ and $k$ with $y$.

This is the equation of a circle centered at the origin with radius $c$. …