Answer:

 Here the given curve

\begin{align} \pmb{ \left(y-x\right)^{2}-a^{2}+x^{2}=0} \end{align}

 (a) To find the whole area :
  Let us first sketch the curve
  Putting \(\pmb{y=0 }\) in (1), we get
\begin{align} &\pmb{ x^{2}-a^{2}+x^{2}=0}\nonumber\\ \implies & \pmb{ 2x^{2}=a^{2}}\nonumber\\ \implies & \pmb{ x=\pm \frac{a}{\sqrt{2}}}\nonumber\\ \end{align}   Therefore the curve intersect with the x-axis at \(\pmb{ A=(\frac{a}{\sqrt{2}},0 )} \) and \( \pmb{ B=(-\frac{a}{\sqrt{2}},0 )}\).
  Also,
\begin{align} &\pmb{ \left(y-x\right)^{2}-a^{2}+x^{2}=0}\nonumber\\ \implies & \pmb{ \left(y-x\right)^{2}=a^{2}-x^{2}}\nonumber\\ \implies & \pmb{ y-x=\pm \sqrt{\left(a^{2}-x^{2} \right)}}\nonumber\\ \implies & \pmb{ y=x \pm \sqrt{\left(a^{2}-x^{2} \right)}}\nonumber\\ \implies & \pmb{ y=f(x)=x + \sqrt{\left(a^{2}-x^{2} \right)}}\\ and~ & \pmb{ y=g(x)=x – \sqrt{\left(a^{2}-x^{2} \right)}} \end{align}   From the above equation, \(\pmb{a^{2}-x^{2} } \) must be non-negative. That is \begin{align} & \pmb{a^{2}-x^{2} \ge 0} \nonumber\\ \implies & \pmb{x^{2}\le a^{2}} \nonumber\\ \implies & \pmb{x\le a}~ and~ \pmb{x\ge -a} \nonumber\\ \end{align}   Putting \( \pmb{x= a} \) in (1), we get
\begin{align} &\pmb{ \left(y-a\right)^{2}-a^{2}+a^{2}=0}\nonumber\\ \implies &\pmb{ \left(y-a\right)^{2}=0}\nonumber\\ \implies &\pmb{ y-a=0}\nonumber\\ \implies &\pmb{ y=a}\nonumber\\ \end{align}   Again putting \( \pmb{x=- a} \) in (1), we get
\begin{align} &\pmb{ \left(y+a\right)^{2}-a^{2}+a^{2}=0}\nonumber\\ \implies &\pmb{ \left(y+a\right)^{2}=0}\nonumber\\ \implies &\pmb{ y+a=0}\nonumber\\ \implies &\pmb{ y=-a} \end{align}   Therefore the points \(\pmb{ C=(a,a )} \) and \(\pmb{ D=(-a,-a )} \) lie on the curve.
  Joining the points C and D we get the chord
\begin{align} &\pmb{ \frac{y-a}{x-a}=\frac{a-(-a)}{a-(-a)}}\nonumber\\ \implies &\pmb{ \frac{y-a}{x-a}=\frac{2a}{2a}}\nonumber\\ \implies &\pmb{ \frac{y-a}{x-a}=1}\nonumber\\ \implies &\pmb{ y-a =x-a}\nonumber\\ \implies &\pmb{ y =h(x)=x} \end{align}   Plotting the curve and the chord, we get

  Therefore the Total Area:

\begin{align} &\pmb{ 2\int^{a}_{-a} \left[f(x)-h(x)\right]dx }\nonumber\\ &=\pmb{ 2\int^{a}_{-a} \left[x + \sqrt{\left(a^{2}-x^{2} \right)}-x\right]dx }\nonumber\\ &=\pmb{ 2\int^{a}_{-a} \sqrt{a^{2}-x^{2} }~dx }\nonumber\\ &=\pmb{ 2\left[\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a} \right]^{a}_{-a} }\nonumber\\ &=\pmb{ 2\left[\frac{a^{2}}{2}\frac{\pi}{2}+\frac{a^{2}}{2}\frac{\pi}{2}\right] }\nonumber\\ &=\pmb{ 2\left[\frac{a^{2}\pi}{2}\right] }\nonumber\\ &=\pmb{ a^{2}\pi }\nonumber\\ \end{align}