If a foot of the normal from the point ( 4 , 3 ) to a circle is ( 2 , 1 ) and 2 x - y - 2 = 0 is a diameter…

Given, the normal from the point $\left(4, 3\right)$ to a circle is $\left(2, 1\right)$.

So, the equation of this normal is $y-1=\frac{3-1}{4-2}\left(x-2\right) \Rightarrow y=x-1 ...\left(1\right)$,

As, the normal at any point to the circle passes through its centre.

And, we have a given diameter $2x-y-2=0 ...\left(2\right)$
Hence, the centre of the circle is the point of intersection of $\left(1\right)$ and $\left(2\right)$.

Solving equations $\left(1\right)$ and $\left(2\right)$, the centre is $\left(1, 0\right)$.

And, the radius is equal to the distance between points $\left(1, 0\right) \mathrm{and} \left(2,1\right)$.

So, $r=\sqrt{{\left(2-1\right)}^2+{\left(1-0\right)}^2}=\sqrt{2}$

Hence, the equation of the circle is 

${\left(x-1\right)}^2+{\left(y-0\right)}^2={\left(\sqrt{2}\right)}^2 \Rightarrow x^2+y^2-2x-1=0$