A right angled triangle $ABC$ having right angle at $C$, $CA = 3 $and $CB = 4$, moves such that angular points $A$ and $B$ slide along x axis and y axis respectively. Find the locus of $C$.
I assumed $C$ to be $(h,k)$ and the points $A$ and $B$ as $(\alpha,0)$ and $(0,\beta)$.
Then as $C$ is right angle,
$\frac{k}{h -\alpha}× \frac{k - \beta}{h} = -1 $ But after this I am stuck I am not able to eliminate $\alpha$ and $\beta$.
Please help me. Thanks in advance.
Hint:
You have $(h-\alpha)^2+k^2=3^2$ and $h^2+(k-\beta)^2=4^2$.
Since $\displaystyle \frac{k}{h-\alpha}\times\frac{k-\beta}{h}=-1$, $h(h-\alpha)=k(k-\beta)$.
\begin{align*} 9h^2-16k^2&=h^2[(h-\alpha)^2+k^2]-k^2[h^2+(k-\beta)^2]\\ 9h^2-16k^2&=0\\ 3h&=\pm4k \end{align*}
Note that $h\in[-4,4]$, $k\in[-3,3]$.