Let the locus of the midpoints of the chords of the circle x^2+(y-1)^2=1 drawn from the or…

Question

Let the locus of the midpoints of the chords of the circle $x^2+(y-1)^2=1$ drawn from the origin intersect the line $x+y=1$ at $\mathrm{P}$ and $\mathrm{Q}$. Then, the length of $\mathrm{PQ}$ is :

Accepted Answer

Let mid-point is $(x,y)$

$\begin{gathered} {x}^{2}+{y}^{2}-2y=0 \ x{x}_{1}+y{y}_{1}-(y+{y}_{1})={x}_{2}^{1}+{y}_{2}^{1}-2{y}_{1} \end{gathered}$

It is passing through origin

$\begin{gathered} 0+0-(0+{y}_{1})={x}_{2}^{1}+{y}_{2}^{1}-2{y}_{1} \ \Rightarrow -{y}_{1}={x}_{2}^{1}+{y}_{2}^{1}-2{y}_{1} \ \Rightarrow {x}_{2}^{1}+{y}_{2}^{1}-{y}_{1}=0 \end{gathered}$

${x}^{2}+{y}^{2}-y=0$ ............ (1)

$\because $ It intersects the line $x+y=1$

So put $x=1-y$ in equation (1)

$\begin{gathered} (1-y{)}^{2}+{y}^{2}-y=0 \ 2{y}^{2}-3y+1=0 \end{gathered}$

$\begin{gathered} \Rightarrow (y-1)(2y-1)=0 \ \Rightarrow y=1,\frac{1}{2} \ 	herefore P(0,1)Q(\frac{1}{2},\frac{1}{2}) \end{gathered}$

So, $PQ=\sqrt{{(\frac{1}{2}-0)}^{2}+{(\frac{1}{2}-1)}^{2}}=\frac{1}{\sqrt{2}}$

Solutions · 1