Question

A triangle has two of its sides along the axes. If the third side touches the circle $x^2+y^2-2ax-2ay+a^2=0$, then the equation of the locus of the circumcenter of the triangle is

• A. $2a(x+y)=2xy+a^2$
• B. $2a(x-y)=2xy+a^2$
• C. $2a(x+y)=2xy-a^2$
• D. None of these

Answer 1

The correct option is A $2a(x+y)=2xy+a^2$

Let the third side be $\frac{x}{\alpha }+\frac{y}{\beta }=1.$

For the circle, centre $\equiv (a,a)$ and radius $=a.$

Since the third side touches the circle,

$\therefore a=\frac{\frac{a}{\alpha }+\frac{a}{\beta }-1}{\sqrt{\frac{1}{a^2}+\frac{1}{{\beta }^2}}}$ ...(1)

Vertices of the triangle are $(0,0),(\alpha ,0)$ and $(0,\beta ),$

$\therefore$ if the circumcentre is $(\gamma ,\delta )$ then

$\gamma =\frac{\alpha }{2}$ and $\delta =\frac{\beta }{2}.$

$\therefore$ From (1), $a^2(\frac{1}{{4r}^2}+\frac{1}{{4\delta }^2})={(\frac{a}{2\gamma }+\frac{a}{2\delta }-1)}^2$

$\Rightarrow 2a(\gamma +\delta )-a^2=2\gamma \delta$

So, the locus of $(\gamma ,\delta )$ is $2a(x+y)=2xy+a^2$