Locus of foot of perpendicular drawn from a fixed point $(10,0)$ on the $x$-axis to any tangent to the circle $x^2+y^2=16$ is
My Attempt
Let the tangent be of slope $m$. So equation of tangent is $y=mx\pm 4\sqrt{1+m^2}$. Now let the coordinates of the foot of perpendicular be $(x_1,y_1)$.
$$\dfrac{x_1-10}{-m}=\dfrac{y_1-0}{1}=-\left(\dfrac{-10m+4\sqrt{1+m^2}}{1+m^2}\right)$$
This gives me the expressions for $x_1$ and $y_1$ as follows from which I have to eliminate the varying parameter $m$.
$$\boxed{x_1=\dfrac{m(4\sqrt{1+m^2}-10m)}{1+m^2}+10} \ \boxed{y_1=\dfrac{10m-4\sqrt{1+m^2}}{1+m^2}}$$
I am not able to proceed on how to eliminate $m$. Any hints are appreciated. Even suggestions to solving this using different methods are welcome. Thanks
Best Answer
Let $L(h,k)$ be the foot of the perpendicular line segment drawn from $P(10,0)$ on the tangent line. Then the slope of this line segment is given by $$m=\frac{k}{h-10}.$$ Let the point of tangency be $Q(x_1,y_1)$. Then the equation of the tangent line at $Q$ is given by $xx_1+yy_1=16$. So the slope of this line is $-\frac{x_1}{y_1}$. Using perpendicularity, we get $$\frac{k}{h-10}=\frac{y_1}{x_1} \implies \color{red}{kx_1+(10-h)y_1=0}.$$
But the point $L$ also lies on it, so $$\color{red}{hx_1+ky_1=16}.$$ Solving for $x_1$ from the first equation , we get $$x_1=\frac{(h-10)y_1}{k}.$$ Substituting this into the second equation we get $$\color{blue}{y_1=\frac{16k}{h(h-10)+k^2}.}$$ Then
$$\color{blue}{x_1=\frac{16(h-10)}{h(h-10)+k^2}.}$$ Now use the fact that $Q(x_1,y_1)$ lies on the circle, to get $$x_1^2+y_1^2=16 \implies \left[\frac{16(h-10)}{h(h-10)+k^2}\right]^2+\left[\frac{16k}{h(h-10)+k^2}\right]^2=16.$$ This simplifies to $$(h(h-10)+k^2)^2=16((h-10)^2+k^2).$$