Question

A straight line passes through a fixed point $(h,k)$. The locus of the foot of perpendicular on it drawn from the origin is

• A. $x^2+y^2–hx–ky=0$
• B. $x^2+y^2+hx+ky=0$
• C. $3x^2+3y^2+hx–ky=0$
• D. None of these

Answer 1

The correct option is A

$x^2+y^2–hx–ky=0$

Explanation for the correct answer:

Step 1: Finding the slope.

The equation of the straight line passes the point $(h,k)$ and the slope $m$ is

$$\begin{gathered} \Rightarrow (y-k)=m(x-h) \\ \Rightarrow y-k=mx-mh \\ \Rightarrow y=mx-mh+k...\left(1\right) \end{gathered}$$

The other line passes perpendicular to the origin $(0,0)$.

Then the slope of this line is $-\frac{1}{m}$

So the equation of perpendicular line is

$$\begin{gathered} y=-\frac{1}{m}x \\ m=-\frac{x}{y}...\left(2\right) \end{gathered}$$

Step 2: Finding the locus

The foot of the perpendicular is the intersection of equation 1 and 2.

The locus can be obtained by

$$\begin{gathered} \Rightarrow y=\left(-\frac{x}{y}\right)x-\left(-\frac{x}{y}\right)h+k \\ \Rightarrow y=-\frac{x^2}{y}+\frac{xh}{y}+k \\ \Rightarrow y^2=-x^2+hx+ky \\ \Rightarrow x^2+y^2-hx-ky=0 \end{gathered}$$

Hence, option (A) ${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{–}\mathit{h}\mathit{x}\mathbf{–}\mathit{k}\mathit{y}\mathbf{=}\mathbf{0}$ is the correct answer.