JEE Main 2024 — Circle Question with Solution

Let the mid point of the chords be $\left(x_1,y_1\right)$.

The given equation of circle is $x^2+{\left(y-1\right)}^2=1$.

$\Rightarrow x^2+y^2-2y=0$

$\Rightarrow x{x}_1+y{y}_1-\left(y+y_1\right)={x_1}^2+{y_1}^2-2y_1$

This locus passes through the origin,

$\Rightarrow 0+0-\left(0+y_1\right)={x_1}^2+{y_1}^2-2y_1$

$\Rightarrow -y_1={x_1}^2+{y_1}^2-2y_1$

$\Rightarrow {x_1}^2+{y_1}^2-y_1=0$

So, the locus is $x^2+y^2-y=0$

It intersects with the line $x+y=1$.

$\Rightarrow {\left(1-y\right)}^2+y^2-y=0$

$\Rightarrow 1+y^2-2y+y^2-y=0$

$\Rightarrow 2y^2-2y-y+1=0$

$\Rightarrow 2y\left(y-1\right)-\left(y-1\right)=0$

$\Rightarrow \left(2y-1\right)\left(y-1\right)=0$

$\Rightarrow y=1,\frac{1}{2}$

$\Rightarrow P\equiv \left(0,1\right) \text{and} Q\left(\frac{1}{2},\frac{1}{2}\right)$

$\Rightarrow PQ=\sqrt{{\left(0-\frac{1}{2}\right)}^2+{\left(1-\frac{1}{2}\right)}^2}$

$\Rightarrow PQ=\sqrt{\frac{1}{4}+\frac{1}{4}}$

$\Rightarrow PQ=\frac{1}{\sqrt{2}}$