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 P and Q . Then, the length of PQ is :

Circle PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q43 with Solution

View written solutionFree

Correct answer: C

Given circle

The circle is $x^2+(y-1)^2=1.$ Its center is $C(0,1)$ and radius is $1$ .

We need the locus of the midpoints of chords of this circle drawn from the origin.

Here, “drawn from the origin” means the chords pass through the origin $O(0,0)$ .

Equation of any chord through the origin

A variable line through the origin can be written as $y=mx.$ This line cuts the circle at two points, one of which is clearly the origin itself because $0^2+(0-1)^2=1,$ so $(0,0)$ lies on the circle.

Thus the chord through the origin joins $O(0,0)$ to the second point of intersection of the line with the circle.

Find the second point of intersection

Substitute $y=mx$ into the circle: $x^2+(mx-1)^2=1.$ Expand: $x^2+m^2x^2-2mx+1=1.$ So, $(1+m^2)x^2-2mx=0.$ Factor: $x\big((1+m^2)x-2m\big)=0.$ Hence the intersections are:

$x=0$ giving the origin,
or $x=\frac{2m}{1+m^2}.$ Then $y=mx=\frac{2m^2}{1+m^2}.$ So the second point is $\left(\frac{2m}{1+m^2},\frac{2m^2}{1+m^2}\right).$

Midpoint of the chord

The chord has endpoints $\left(0,0\right),\quad \left(\frac{2m}{1+m^2},\frac{2m^2}{1+m^2}\right).$

Its midpoint is $M\left(\frac{m}{1+m^2},\frac{m^2}{1+m^2}\right).$

Let $M=(h,k)$ . Then $h=\frac{m}{1+m^2},\qquad k=\frac{m^2}{1+m^2}.$

Now eliminate $m$ .

From these, $\frac{h}{k}=\frac{m}{m^2}=\frac{1}{m} \quad (m\neq 0),$ so $m=\frac{k}{h}$ , but there is a simpler relation:

Since $k=\frac{m^2}{1+m^2},\qquad h=\frac{m}{1+m^2},$ we get $h^2=\frac{m^2}{(1+m^2)^2}.$ Also,

=\frac{m^2(1+m^2)-m^4}{(1+m^2)^2} =\frac{m^2}{(1+m^2)^2}=h^2.$$ Thus, $$h^2=k-k^2.$$ So, $$h^2+k^2-k=0.$$ Replacing $(h,k)$ by $(x,y)$, the locus is $$x^2+y^2-y=0,$$ or $$x^2+\left(y-\frac12\right)^2=\left(\frac12\right)^2.$$ So the locus is a circle with center $\left(0,\frac12\right)$ and radius $\frac12$. --- 5. **Intersect this locus with the line $x+y=1$** From the line, $$y=1-x.$$ Substitute into the locus: $$x^2+(1-x)^2-(1-x)=0.$$ Expand: $$x^2+1-2x+x^2-1+x=0,$$ $$2x^2-x=0,$$ $$x(2x-1)=0.$$ Hence $$x=0 \quad \text{or} \quad x=\frac12.$$ Then corresponding $y$ values are - if $x=0$, $y=1$, - if $x=\frac12$, $y=\frac12$. Thus the intersection points are $$P=(0,1),\qquad Q=\left(\frac12,\frac12\right).$$ --- 6. **Find $PQ$** $$PQ=\sqrt{\left(\frac12-0\right)^2+\left(\frac12-1\right)^2} =\sqrt{\frac14+\frac14} =\sqrt{\frac12} =\frac{1}{\sqrt{2}}.$$ --- 7. **Final answer** $$\boxed{\frac{1}{\sqrt{2}}}$$ So the correct option is **C**.

More from Circle

View written solutionFree

Correct answer: C

Given circle

The circle is $x^2+(y-1)^2=1.$ Its center is $C(0,1)$ and radius is $1$ .

We need the locus of the midpoints of chords of this circle drawn from the origin.

Here, “drawn from the origin” means the chords pass through the origin $O(0,0)$ .

Equation of any chord through the origin

A variable line through the origin can be written as $y=mx.$ This line cuts the circle at two points, one of which is clearly the origin itself because $0^2+(0-1)^2=1,$ so $(0,0)$ lies on the circle.

Thus the chord through the origin joins $O(0,0)$ to the second point of intersection of the line with the circle.

Find the second point of intersection

Substitute $y=mx$ into the circle: $x^2+(mx-1)^2=1.$ Expand: $x^2+m^2x^2-2mx+1=1.$ So, $(1+m^2)x^2-2mx=0.$ Factor: $x\big((1+m^2)x-2m\big)=0.$ Hence the intersections are:

$x=0$ giving the origin,
or $x=\frac{2m}{1+m^2}.$ Then $y=mx=\frac{2m^2}{1+m^2}.$ So the second point is $\left(\frac{2m}{1+m^2},\frac{2m^2}{1+m^2}\right).$

Midpoint of the chord

The chord has endpoints $\left(0,0\right),\quad \left(\frac{2m}{1+m^2},\frac{2m^2}{1+m^2}\right).$

Its midpoint is $M\left(\frac{m}{1+m^2},\frac{m^2}{1+m^2}\right).$

Let $M=(h,k)$ . Then $h=\frac{m}{1+m^2},\qquad k=\frac{m^2}{1+m^2}.$

Now eliminate $m$ .

From these, $\frac{h}{k}=\frac{m}{m^2}=\frac{1}{m} \quad (m\neq 0),$ so $m=\frac{k}{h}$ , but there is a simpler relation:

Since $k=\frac{m^2}{1+m^2},\qquad h=\frac{m}{1+m^2},$ we get $h^2=\frac{m^2}{(1+m^2)^2}.$ Also,

=\frac{m^2(1+m^2)-m^4}{(1+m^2)^2} =\frac{m^2}{(1+m^2)^2}=h^2.$$ Thus, $$h^2=k-k^2.$$ So, $$h^2+k^2-k=0.$$ Replacing $(h,k)$ by $(x,y)$, the locus is $$x^2+y^2-y=0,$$ or $$x^2+\left(y-\frac12\right)^2=\left(\frac12\right)^2.$$ So the locus is a circle with center $\left(0,\frac12\right)$ and radius $\frac12$. --- 5. **Intersect this locus with the line $x+y=1$** From the line, $$y=1-x.$$ Substitute into the locus: $$x^2+(1-x)^2-(1-x)=0.$$ Expand: $$x^2+1-2x+x^2-1+x=0,$$ $$2x^2-x=0,$$ $$x(2x-1)=0.$$ Hence $$x=0 \quad \text{or} \quad x=\frac12.$$ Then corresponding $y$ values are - if $x=0$, $y=1$, - if $x=\frac12$, $y=\frac12$. Thus the intersection points are $$P=(0,1),\qquad Q=\left(\frac12,\frac12\right).$$ --- 6. **Find $PQ$** $$PQ=\sqrt{\left(\frac12-0\right)^2+\left(\frac12-1\right)^2} =\sqrt{\frac14+\frac14} =\sqrt{\frac12} =\frac{1}{\sqrt{2}}.$$ --- 7. **Final answer** $$\boxed{\frac{1}{\sqrt{2}}}$$ So the correct option is **C**.

Circle question

Circle question

More from Circle

Circle question

Circle question

More from Circle