Question

Find the locus of the circumcenter of the triangle whose two sides are along the co-ordinate axes and third side passes through the point of intersection of the lines $ax+by+c=0$ and $lx+my+n=0$

Answer 1

As two sides are along coordinate axes, hence triangle will be right angled.

Let locus of circumcentre be $(h,k)$ and equation of hypotenuse be $\frac{x}{p}+\frac{y}{q}=1$.

As triangle is right angled hence circumcentre will be at the mid point of the hypotenuse.

$\Rightarrow (h,k)=(\frac{p+0}{2},\frac{0+q}{2})$

$\Rightarrow p=2h$ and $q=2k$

$\Rightarrow \frac{x}{p}+\frac{y}{q}=1$

$\Rightarrow \frac{x}{2h}+\frac{y}{2k}=1$

$\Rightarrow \frac{x}{h}+\frac{y}{k}=2$ ….$\left(1\right)$

Given that hypotenuse passes through $ax+by+c=0$ and $lx+my+n=0$

On solving we get,

$\Rightarrow x=\frac{bn-cm}{am-bl}$ and $y=\frac{an-cl}{bl-am}$

Putting the value of $x,y$ in equation $\left(1\right)$, we get,

$\Rightarrow \frac{bn-cm}{am-bl}\frac{1}{h}+\frac{an-cl}{bl-am}\frac{1}{k}=2$

$\Rightarrow k(bn-cm)(bl-am)+h(an-cl)(am-bl)=2hk(am-bl)(bl-am)$

$\Rightarrow$ Replace $h\to x$ and $k\to y$ to get the condition of locus of circumcenter.

$\Rightarrow y(bn-cm)(bl-am)+x(an-cl)(am-bl)=2xy(am-bl)(bl-am)$