Question

A circle of constant radius 5 units passes through the origin $O$ and cuts the axes at $A$ and $B$. Then the locus of the foot of the perpendicular from $O$ to $AB$ is $(x^2 + y^2)^2(\frac{1}{x^2}+\frac{1}{y^2}) = k$, then $k$ is equal to

Answer 1

100

Answer 2

The general equation of a circle passing through the origin is $x^2 + y^2 + 2gx + 2fy = 0$. The radius is given as 5, so $g^2+f^2 = 5^2 = 25$. The circle cuts the x-axis at $A(-2g, 0)$ and the y-axis at $B(0, -2f)$. The equation of line AB is $\frac{x}{-2g} + \frac{y}{-2f} = 1$, which simplifies to $fx + gy + 2fg = 0$. Let $(h, k)$ be the foot of the perpendicular from $O(0,0)$ to AB. The line OR (where R is $(h, k)$) is perpendicular to AB. The slope of AB is $-f/g$. The slope of OR is $k/h$. Since OR $\perp$ AB, $(k/h)(-f/g) = -1$, so $kf = hg$. The distance $OR = \frac{|f(0) + g(0) + 2fg|}{\sqrt{f^2+g^2}} = \frac{|2fg|}{5}$. Also, $OR^2 = h^2+k^2$. So, $h^2+k^2 = \frac{4f^2g^2}{25}$, which means $25(h^2+k^2) = 4f^2g^2$. From $kf=hg$, we have $f^2g^2 = k^2f^2 \cdot \frac{g^2}{f^2} = k^2f^2 \cdot \frac{h^2}{k^2} = h^2k^2$. Substituting $f^2g^2 = h^2k^2$ into $25(h^2+k^2) = 4f^2g^2$: $25(h^2+k^2) = 4h^2k^2$. This step seems to be wrong in the explanation. Let's correct it.

From $kf=hg$, we have $f^2 = \frac{h^2g^2}{k^2}$. Substitute this into $g^2+f^2=25$: $g^2 + \frac{h^2g^2}{k^2} = 25 \implies g^2(\frac{k^2+h^2}{k^2}) = 25 \implies g^2 = \frac{25k^2}{h^2+k^2}$. Similarly, from $g^2 = \frac{k^2f^2}{h^2}$, substitute into $g^2+f^2=25$: $\frac{k^2f^2}{h^2} + f^2 = 25 \implies f^2(\frac{k^2+h^2}{h^2}) = 25 \implies f^2 = \frac{25h^2}{h^2+k^2}$.

Now substitute $f^2$ and $g^2$ into $25(h^2+k^2) = 4f^2g^2$: $25(h^2+k^2) = 4 \left(\frac{25h^2}{h^2+k^2}\right) \left(\frac{25k^2}{h^2+k^2}\right)$ $25(h^2+k^2) = \frac{2500h^2k^2}{(h^2+k^2)^2}$ $(h^2+k^2)^3 = 100h^2k^2$. Dividing by $h^2k^2$: $\frac{(h^2+k^2)^3}{h^2k^2} = 100$ $(h^2+k^2)^2 \left(\frac{h^2+k^2}{h^2k^2}\right) = 100$ $(h^2+k^2)^2 \left(\frac{1}{k^2}+\frac{1}{h^2}\right) = 100$. Replacing $(h,k)$ with $(x,y)$: $(x^2+y^2)^2(\frac{1}{x^2}+\frac{1}{y^2}) = 100$. Comparing with the given equation, $k=100$.