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Question

If the chord of contact of tangents drawn from a point (α,β) to the circle x2+y2=a2 subtends a right angle at the centre of the circle, then:

A
α2+β2=a2
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B
α2+β2=2a2
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C
α2+β2=2/a2
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D
α2+β2=a2/2
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Solution

The correct option is C α2+β2=2a2
The given equation of circle is,
x2+y2=a2
(x0)2+(y0)2=a2

Comparing this equation with standard form i.e. (xh)2+(yk)2=r2, we get
h=0, k=0 and r=a

Thus, Center of circle is at origin i.e. O(0,0) and OA=OB=a

Now, AC is tangent to the circle through point A.
CAO=900

Similarly, BC is tangent to circle to a circle through point B.
CBO=900

Also, AOB=900 (Given)
Thus, we can conclude that ACB=900

Similarly, OA=OB=AC=BC
Thus, AOBC is square.

Now, by distance formula, OC=(α0)2+(β0)2
OC=α2+β2

By property of square, diagonals are equal in length.
AB=OC=α2+β2

Now, in right angled triangle ΔAOB, by Pythagoras theorem,
AB2=OA2+OB2
(α2+β2)2=a2+a2
α2+β2=2a2

Thus, answer is option (B)




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