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Question

The locus of a point, which is such that the length of the tangents from its two concentric circles of radii $$a$$ and $$b$$ are inversely proportional to their radii, is a circle $$C$$ of area $$16 \pi$$


A
centre of C is at the centre of the given circles
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B
centre of C is at a distance 4 from the centre of the given circles
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C
a2+b2=16
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D
a2+b2=4
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Solution

The correct options are
A centre of $$C$$ is at the centre of the given circles
B $$\displaystyle a^{2}+b^{2}=16$$
Let $$a>b$$
Let the first circles be $$x^2 +y^2 = a^2$$ and $$x^2+y^2 = b^2 $$. 
Let $$t_1 $$ and $$t_2$$ be the length of the tangents drawn from $$P(h,k)$$. 
As per the given conditions, 
$$t_1 \times a = t_2 \times b$$  -- (i)
By Pythagoras theorem, 
$$t_1^2 = h^2+k^2 - a^2 $$
$$t_2^2 = h^2+k^2 -b^2 $$
Squaring (i) , 
$$t_1^2 a^2 = t_2^2b^2 $$
$$(h^2+k^2-a^2)a^2 = (h^2+k^2-b^2)b^2 $$
$$(h^2+k^2) (a^2 -b^2) = a^4-b^4$$
$$h^2+k^2 = a^2+b^2 $$
Since the area is 16$$\pi$$, the radius is equal to $$4$$
Hence, 
$$a^2+b^2 =16$$
Also, C is the centre of the circle
Hence, options A and C are correct.

Mathematics

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