If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then \(a\) is equal to

Given curves are,
\(x^{2}+y^{2}=4 x...(i)\)
On differentiating w.r.t. \(x\), we get
\(2 x+2 y \frac{d y}{d x}=4\)
\(\Rightarrow \quad \frac{d y}{d x}=\frac{2-x}{y}\)
At \((2,2), \quad\left(\frac{d y}{d x}\right)_{(2,2)}=\frac{2-0}{2}=0=m_{1}\)
and \(\quad x^{2}+y^{2}=8\)
On differentiating w.r.t. \(x\), we get
\(\begin{aligned}
2 x+2 y \frac{d y}{d x} &=0 \\
\Rightarrow \quad \frac{d y}{d x} &=\frac{-x}{y}
\end{aligned}\)
At \((2,2)\)
\(\left(\frac{d y}{d x}\right)_{(2,2)}=\frac{-2}{2}=-1=m_{2}\)
Let \(\theta\) be an acute angle between the given curves, then, we get
\(\begin{array}{cc}
& \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} \cdot m_{2}}\right|=\left|\frac{0+1}{1+0}\right|=1 \\
\Rightarrow & \tan \theta=\tan 45^{\circ} \\
\Rightarrow & \theta=45^{\circ}
\end{array}\)
\(\Rightarrow \quad \sin ^{-1}(a)=45^{\circ} \quad\) (given \(\left.\theta=\sin ^{-1} a\right)\)
\(\Rightarrow \quad a=\sin 45^{\circ}\)
\(\therefore \quad a=\frac{1}{\sqrt{2}}\)