Question

The locus of midpoints of the chords of contact of $x^2+y^2=2$ from the points on the line $3x+4y=10$ is a circle with centre $P$. If $O$ be the origin, then $OP$ is equal to

• A. $2$
• B. $3$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$\frac{1}{2}$

Let $M$$(h,k)$ be the midpoint of the chord of contact.
Now, the line joining point $M$ and the origin is perpendicular to the chord of contact.
Hence, the slope of the line $=-\frac{h}{k}$.
The equation of the chord of contact will be $hx+ky=h^2+k^2$ ..... (i)
The equation of the chord of contact can also be written as $(x_1,y_1)$ is $x{x}_1+y{y}_1=2$ ...... (ii)
Comparing the two equations, we get

$x_1=\frac{2h}{h^2+k^2}$ and $y_1=\frac{2k}{h^2+k^2}$
$(x_1,y_1)$ lies on $3x+4y=10\Rightarrow 6h+8k=10(h^2+k^2)$
$\therefore$ Locus of $(h,k)$ is $x^2+y^2-\frac{3}{5}x-\frac{4}{5}y=0$ which is circle with centre $P$ $(\frac{3}{10},\frac{4}{10})$.
$\therefore OP=\frac{1}{2}$