The correct option is
C α2+β2=2a2The given equation of circle is,
x2+y2=a2
∴(x−0)2+(y−0)2=a2
Comparing this equation with standard form i.e. (x−h)2+(y−k)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=√(α−0)2+(β−0)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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