Question

Tangents PA and PB are drawn from a point P to the circle $x^2+y^2-2x-2y+1=0$. If the point P lies on the line $lx+my+n=0$, where l, m, n are constants, then find the locus of the circum-centre of the $\Delta$PAB.

Answer 1

Here, we see the formation of two right angled triangles $△POA$ & $△POB$. We see that $△POA$ & $△POB$ share a hypotenuse $OP$.

We also know that circumcentre of $△POA$ & $△OPB$ will lie at the midpoint of hypotenuse $OP$.

$\therefore$ The $△POA$ & $△POB$ share a circumcircle. This circumcircle passes through $P,A,O$ & $B$ $\therefore$ It is also circumcircle of $△PAB$

$\therefore$ Circumcentre of $△PAB\equiv (h,k)\equiv (\frac{x+1}{2},\frac{y+1}{2})$

$\Rightarrow x=2h-1$ & $y=2k-1$

As $(x,y)$ lies of line

$lx+my+n=0$

$\Rightarrow l[2h-1]+m[2k-1]+n=0$

$\Rightarrow 2lh+2mk=l+m-n$

$\therefore Locusof=circumcentre=2lx+2my-l-m+n=0$