Question

Through the origin, chords are drawn to the circle ${(x-1)}^2+y^2=1$. Find the equation of the locus of the mid points of these chords.

Answer 1

We have,

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

Now,

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

On comparing that,

${(x-h)}^2+{(y-k)}^2=r^2$

So,

$(h,k)=(1,0)$

The mid point of chord is $(h_1,k_1)=(\frac{0+1}{2},\frac{0+0}{2})=(\frac{1}{2},0)$

Now, equation of chord is

${(x-h_1)}^2+{(y-k_1)}^2=r^2$

$\Rightarrow {(x-\frac{1}{2})}^2+{(y-0)}^2=1^2$

$\Rightarrow x^2+\frac{1}{4}-x+y^2=1$

$\Rightarrow x^2+y^2-x=1-\frac{1}{4}$

$\Rightarrow x^2+y^2-x=\frac{3}{4}$

Hence, this is the answer.