Question

Consider the circle $c:x^2+y^2=1$ and the line $L:y=m(x+2)$. If line L intersect c at P and Q, then locus of middle point of PQ is

• A. $(x+1{)}^2+y^2=1$
• B. $x^2+(y-1{)}^2=1$
• C. $(x-1{)}^2+y^2=1$
• D. $x^2+(y+1{)}^2=1$

Answer 1

The correct option is A $(x+1{)}^2+y^2=1$

Using $T=S_1$ take mid – point (h, k) then equation of PQ is
$xh+yk-(h^2+k^2)=0\dots \dots \left(1\right)$
$mx-y+2m=0\dots \dots \left(2\right)$
Compare (1) and (2) eliminate m