Question

A variable straight line through $A(-1,-1)$ is drawn to cut the circle $x^2+y^2=1$ at the points $B,C.$ If $P$ is chosen on the line $ABC$ such that $AB,AP,AC$ are in $H.P$ then the locus of $P$ is

• A. $x+y+1=0$
• B. $x+y-1=0$
• C. $x-y+1=0$
• D. $x-y-1=0$

Answer 1

The correct option is A $x+y+1=0$

Let locus at point $P=(h,k),AP=r,$ inclination of line $ABC$ be $\theta$ then any point on the line will be of form $(x,y)=(-1+r\cos \theta ,-1+r\sin \theta )$
For this point to lie on the circle $x^2+y^2=1$
$\Rightarrow (-1+r\cos \theta {)}^2+(-1+r\sin \theta {)}^2=1$
$\Rightarrow r^2-2(\cos \theta +\sin \theta )r+1=0$
$r_1,r_2$ are its roots
Let $r_1=AB,r_2=AC$
If $AB,AP,AC$ are in $H.P$
$\Rightarrow AP=\frac{2AB.AC}{AB+AC}$
$\Rightarrow r=\frac{2r_1\cdot r_2}{r_1+r_2}$
$\Rightarrow r=\frac{2}{2(\cos \theta +\sin \theta )}$
$\Rightarrow r\cos \theta +r\sin \theta =1$
$\Rightarrow (h+1)+(k+1)=1$
$\Rightarrow h+k+1=0$
so, locus is $x+y+1=0$