Question

A circle of constant radius 'a' passes through origin 'O' and cuts the axes of co-ordinates in points P and Q, then the equation of the locus of the foot of perpendicular from O to PQ is:

• A. $(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$
• B. $(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2})=a^2$
• C. $(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$
• D. $(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2})=a^2$

Answer 1

Answer: (A), (C)

$(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2})=4a^2$

Answer 2

Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy = 0$ as it passes through the origin. The circle cuts the x-axis at P. Setting $y=0$, we get $x^2 + 2gx = 0$, so $x(x+2g)=0$. Thus, $P = (-2g, 0)$. The circle cuts the y-axis at Q. Setting $x=0$, we get $y^2 + 2fy = 0$, so $y(y+2f)=0$. Thus, $Q = (0, -2f)$. The radius of the circle is given as 'a'. The radius of $x^2 + y^2 + 2gx + 2fy + c = 0$ is $\sqrt{g^2+f^2-c}$. Here $c=0$, so $a = \sqrt{g^2+f^2}$, which means $g^2+f^2 = a^2$. The line PQ passes through $P(-2g, 0)$ and $Q(0, -2f)$. The equation of line PQ is $\frac{x}{-2g} + \frac{y}{-2f} = 1$, which simplifies to $fx + gy + 2fg = 0$. Let R(h, k) be the foot of the perpendicular from the origin O(0,0) to PQ. The slope of PQ is $m_{PQ} = \frac{-2f - 0}{0 - (-2g)} = -\frac{f}{g}$. The slope of OR is $m_{OR} = \frac{k}{h}$. Since OR is perpendicular to PQ, $m_{OR} \cdot m_{PQ} = -1$, so $\frac{k}{h} \cdot (-\frac{f}{g}) = -1$, which gives $kf = hg$. The point R(h, k) also lies on the line PQ, so $fh + gk + 2fg = 0$. From $kf = hg$, we have $g = \frac{kf}{h}$. Substituting this into $fh + gk + 2fg = 0$: $fh + k(\frac{kf}{h}) + 2f(\frac{kf}{h}) = 0$ $fh^2 + k^2f + 2kf^2 = 0$ $f(h^2 + k^2 + 2kf) = 0$. This implies $f=0$ or $h^2 + k^2 + 2kf = 0$. If $f=0$, then $g^2=a^2$. From $kf=hg$, $hg=0$. If $h \neq 0$, then $k=0$, giving R(h,0). If $k \neq 0$, then $h=0$, giving R(0,k). If $f=0$, the line PQ is $gx+2fg=0$, which is $gx=0$ if $g \neq 0$. This means $x=0$. The foot of the perpendicular from origin to $x=0$ is (0,0). If $h^2 + k^2 + 2kf = 0$, then $f = -\frac{h^2+k^2}{2k}$. From $kf=hg$, $g = \frac{kf}{h} = \frac{k}{h} \left(-\frac{h^2+k^2}{2k}\right) = -\frac{h^2+k^2}{2h}$. Substitute these into $g^2+f^2=a^2$: $\left(-\frac{h^2+k^2}{2h}\right)^2 + \left(-\frac{h^2+k^2}{2k}\right)^2 = a^2$ $\frac{(h^2+k^2)^2}{4h^2} + \frac{(h^2+k^2)^2}{4k^2} = a^2$ $(h^2+k^2)^2 \left(\frac{1}{4h^2} + \frac{1}{4k^2}\right) = a^2$ $(h^2+k^2)^2 \left(\frac{k^2+h^2}{4h^2k^2}\right) = a^2$ $\frac{(h^2+k^2)^3}{4h^2k^2} = a^2$ $(h^2+k^2)^3 = 4a^2h^2k^2$. Replacing (h,k) with (x,y): $(x^2+y^2)^3 = 4a^2x^2y^2$. Dividing by $x^2y^2$: $\frac{(x^2+y^2)^3}{x^2y^2} = 4a^2$. $(x^2+y^2)^2 \left(\frac{x^2+y^2}{x^2y^2}\right) = 4a^2$. $(x^2+y^2)^2 \left(\frac{1}{y^2} + \frac{1}{x^2}\right) = 4a^2$.