Single Choice

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$$\dfrac{1}{2}$$
Correct Answer
D$$\dfrac{1}{3}$$

Solution

Let $$ M$$(h,k)$$ be the midpoint of the chord of contact.
Now, the line joining point $$M$$ and the origin is perpendicular to the chord of contact.
Hence, the slope of the line $$ = -\dfrac{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)$$ is $$xx_1 + yy_1 = 2$$ ...... (ii)
Comparing the two equations, we get
$$\displaystyle x_1 = \frac{2h}{h^2 + k^2} $$ and $$y_1 = \displaystyle\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 - \dfrac{3}{5} x - \dfrac{4}{5} y = 0$$ which is circle with centre $$P$$ $$\displaystyle \left ( \dfrac{3}{10}, \dfrac{4}{10} \right )$$.
$$\therefore OP = \dfrac{1}{2}$$

SIMILAR QUESTIONS

Circles

If a circle passing through the point $$(-1, 0)$$ touches y-axis at $$(0, 2)$$, then the length of the chord of the circle along the x-axis is

Circles

Show that the equation of a straight line meeting the circle $$x^2+y^2=a^2$$ in two points at equal distance 'd' from a point $$(x_1, y_1)$$ on its circumference is $$xx_1+yy_1-a^2+\left(\dfrac{d^2}{2}\right)=0$$.

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