Let the locus of the mid points of the chords of circle x^2+(y-1)^2=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of…

Let the mid point of the chords be $({x}_{1},{y}_{1})$(x1​,y1​).

The given equation of circle is ${x}^{2}+{(y-1)}^{2}=1$x2+(y−1)2=1.

$$\Rightarrow {x}^{2}+{y}^{2}-2y=0$$⇒x2+y2−2y=0

$$\Rightarrow x{x}_{1}+y{y}_{1}-(y+{y}_{1})={{x}_{1}}^{2}+{{y}_{1}}^{2}-2{y}_{1}$$⇒xx1​+yy1​−(y+y1​)=x1​2+y1​2−2y1​

This locus passes through the origin,

$$\Rightarrow 0+0-(0+{y}_{1})={{x}_{1}}^{2}+{{y}_{1}}^{2}-2{y}_{1}$$⇒0+0−(0+y1​)=x1​2+y1​2−2y1​

$$\Rightarrow -{y}_{1}={{x}_{1}}^{2}+{{y}_{1}}^{2}-2{y}_{1}$$⇒−y1​=x1​2+y1​2−2y1​

$$\Rightarrow {{x}_{1}}^{2}+{{y}_{1}}^{2}-{y}_{1}=0$$⇒x1​2+y1​2−y1​=0

So, the locus is ${x}^{2}+{y}^{2}-y=0$x2+y2−y=0

It intersects with the line $x+y=1$x+y=1.

$$\Rightarrow {(1-y)}^{2}+{y}^{2}-y=0$$⇒(1−y)2+y2−y=0

$$\Rightarrow 1+{y}^{2}-2y+{y}^{2}-y=0$$⇒1+y2−2y+y2−y=0

$$\Rightarrow 2{y}^{2}-2y-y+1=0$$⇒2y2−2y−y+1=0

$$\Rightarrow 2y(y-1)-(y-1)=0$$⇒2y(y−1)−(y−1)=0

$$\Rightarrow (2y-1)(y-1)=0$$⇒(2y−1)(y−1)=0

$$\Rightarrow y=1,\frac{1}{2}$$⇒y=1,21​

$$\Rightarrow P\equiv (0,1)\text{and}Q(\frac{1}{2},\frac{1}{2})$$⇒P≡(0,1)andQ(21​,21​)

$$\Rightarrow PQ=\sqrt{{(0-\frac{1}{2})}^{2}+{(1-\frac{1}{2})}^{2}}$$⇒PQ=(0−21​)2+(1−21​)2​

$$\Rightarrow PQ=\sqrt{\frac{1}{4}+\frac{1}{4}}$$⇒PQ=41​+41​​

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