Let $C$ be the circle with centre $(0, 0)$ and radius $3$ units. Locus of the point $P$ from which the chord of contact subtends an angle of $60^{\circ}$ at any point on the circumference of $C$ is

x^{2} + y^{2} = 27

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x^{2} + y^{2} = 18

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x^{2} + y^{2} = 36

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x^{2} + y^{2} = 16

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Solution

The correct option is C $x^{2} + y^{2} = 36$
$C : x^{2} + y^{2} = 9$

In triangle $O A B$ ,
$A B = (2 sin 60^{\circ}) O A$

$\Rightarrow \frac{2 (\sqrt{h^{2} + k^{2} - 9}) \cdot 3}{\sqrt{h^{2} + k^{2}}} = \frac{3 \cdot 2 \cdot \sqrt{3}}{2} [From △ O A P]$
$\Rightarrow 4 (h^{2} + k^{2} - 9) = 3 (h^{2} + k^{2})$
$\Rightarrow h^{2} + k^{2} = 36$

Hence, locus is $x^{2} + y^{2} = 36$ .

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Let C be the circle with centre (0,0) and radius 3 units. Locus of the point P from which the chord of contact subtends an angle of 60∘ at any point on the circumference of C is

The correct option is C x2+y2=36C:x2+y2=9 In triangle OAB, AB=(2sin60∘)OA ⇒2(√h2+k2−9)⋅3√h2+k2=3⋅2⋅√32 [From △OAP] ⇒4(h2+k2−9)=3(h2+k2) ⇒h2+k2=36 Hence, locus is x2+y2=36.

Let $C$ be the circle with centre $(0, 0)$ and radius $3$ units. Locus of the point $P$ from which the chord of contact subtends an angle of $60^{\circ}$ at any point on the circumference of $C$ is