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The equation of circle c_{1} is \mathrm{x^2+y^2=16} . The locus of the intersection of orthogonal tangents to the circleis the curve \mathrm{c_{2}} & the locus of the intersection of \mathrm{\perp } tangents to the curve \mathrm{c_{2}} is the curve \mathrm{c_{3}}, then

Option: 1

c_{2} & c_{3} are circles having same centre


Option: 2

The area enclosed by the curve \mathrm{\mathrm{C}_3} is \mathrm{64\pi} sq. unit


Option: 3

The area enclosed by the curve \mathrm{\mathrm{C}_2} is \mathrm{32\pi} sq.unit


Option: 4

All of the above


Answers (1)

best_answer

The locus of the point of intersection of two
perpendicular (orthogonal) tangents to a given conic
is known as its director circle and the equation of the director of the circle \mathrm{x^2+y^2=a^2} is given by \mathrm{x^2+y^2=2a^{2}}

Now,\mathrm{c_2}  is locus of point of intersection of two orthogonal
tangents is director circle of \mathrm{x^{2}+y^{2}=16}, is given by 

\mathrm{x^2+y^2=2.16 \text { i.e., } x^2+y^2=32}

Also the curve \mathrm{\mathrm{C}_3}  is director circle of \mathrm{\mathrm{c}_2} whose equation is given by \mathrm{x^2+y^2=2 \cdot(32) \text { i.e., } x^2+y^2=64}

Thus \mathrm{c_2: x^2+y^2=32 \text { and } c_3: x^2+y^2=64}

So, \mathrm{c_2, c_3} are circles with same centre (0, 0)

Now, \mathrm{c_2: x^2+y^2=32} which is a circle of radius \mathrm{r=\sqrt{32}}

\mathrm{\therefore } Area of curve \mathrm{c_2} is \mathrm{\pi r^2=\pi(32)=32 \pi \text { sq. units }}

again \mathrm{c_3: x^2+y^2=8^2} which is a circle of radius r = 8 

\mathrm{\therefore } Area of curve \mathrm{c_{3} } is \mathrm{\pi r^2=\pi(8)^2=64 \pi \text { sq. units. } }

 

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Rishi

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