Question

The number of ordered real pairs $(x,y)$ with $0<x,y<1$ satisfying the simultaneous equation
$\frac{1+\sqrt{1-x^2}}{x}=1+\frac{2y}{1-y+\sqrt{1-y^2}}$ and
$25(1-x^2)=17-10\sqrt{1-y^2}$
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Answer 1

$\frac{1+\sqrt{1-x^2}}{x}=1+\frac{2y}{1-y+\sqrt{1-y^2}}...\left(1\right)$
$25(1-x^2)=17-10\sqrt{1-y^2}...\left(2\right)$
Since, $0<x,y<1$
Let $x=\sin A,y=\sin B$

$\Rightarrow \frac{1+\cos A}{\sin A}=1+\frac{2\sin B}{1-\sin B+\cos B}$
$\Rightarrow \cot \left(A/2\right)=\frac{1+\cos B+\sin B}{1+\cos B-\sin B}$
$\Rightarrow \cot \left(A/2\right)=\frac{1+\frac{\sin B}{1+\cos B}}{1-\frac{\sin B}{1+\cos B}}$
$\Rightarrow \cot \left(A/2\right)=\frac{1+\tan \left(B/2\right)}{1-\tan \left(B/2\right)}$
$\Rightarrow \tan (\frac{\pi }{2}-\frac{A}{2})=\tan (\frac{\pi }{4}+\frac{B}{2})$
$\Rightarrow \frac{\pi }{2}-\frac{A}{2}=\frac{\pi }{4}+\frac{B}{2}\Rightarrow A+B=\frac{\pi }{2}$

$\therefore \sin (\frac{\pi }{2}-A)=\cos A$
$\Rightarrow x^2+y^2=1$
From eqn (2)
$25(1-x^2)=17-10x\Rightarrow 25x^2-10x-8=0\Rightarrow x=\frac{4}{5},y=\frac{3}{5}(\because 0<x,y<1)$
Only one solution.