Question

$A$ triangle has two of its sides along the axes, its third side touches the circle $x^2+y^2-2ax-2ay+a^2=0$, where $a>0$. If the locus of the circumcentre of the triangle passes through the point $(38,-37)$, then $a^2-2a$ is equal to

Answer 1

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 $\frac{x}{p}+\frac{y}{q}=1$
So that $A$ is $(p,0)$ and $B$ is $(0,q)$ and the line $AB$ touches the given circle.
Since $\angle AOB$ 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, $\frac{a(p+q)-pq}{\sqrt{p^2+q^2}}=a$
$\Rightarrow a^2(p+q{)}^2+p^2{q}^2-2apq(p+q)=a^{,2}(p^2+q^2)$
$\Rightarrow 2a^2-2a(p+q)+pq=0$
$\Rightarrow 2a^2-2a(2h+2k)+2h\cdot 2k=0$.

Hence, the locus of $P(h,k)$ is $a^2-2a(x+y)+2xy=0$.
Since it passes through $(38,-37)$
$a^2-2a=2\times 38\times 37=2812$