A triangle has two of its sides along the axes, its third side touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{ax}-2\mathrm{ay}+{\mathrm{a}}^2=0.$ If the locus of the circumcentre of the triangle passes (38, –37) then a² – 2a is equal to K then the digit in ten’s place is____

The given circle has its centre at C(a,a) and its radius is a, so that it touches both the axes along which lie the two sides of the triangle. 

Let third side be px​+qy​=1
So that A is (p,0) and B is (0,q) and the line AB touches the given circle. 
Since ∠AOB is a right angle, AB is a diameter of the circumcircle of the triangle AOB. 
So, the circumcentre P(h,k) of the triangle AOB is the mid point of AB.
i.e. 2h=p,2k=q 
Now, the equation of AB is,$\frac{x}{p}+\frac{y}{q}=1$ which touches the given circle.
Therefore,$$\begin{gathered} \frac{a(p+q)-pq}{\sqrt{p^2+q^2}}=a \\ \Rightarrow a^2{(p+q)}^2+p^2{q}^2-2apq(p+q)=a^2(p^2+q^2) \\ \Rightarrow 2a^2-2a(p+q)+pq=0 \\ \Rightarrow 2a^2-2a(2h+2k)+2h.2k=0 \\ hence locus of P(h,k) is a^2-2a(x+y)+2xy=0 \end{gathered}$$

Since it passes through (38,−37)
$a^2$−2a=2×38×37=2812