The condition that the chord x cos alpha + y sin alpha  p = 0 of x^2 + y^2  a^2 = 0 may subtend a right angle at the center of the circle is: A (a^2 = 2p^2) B (p^2 = 2a^2) C (a = 2p) D (p = 2a)
Question
The condition that the chord x cos α + y sin α  p = 0 of x^2 + y^2  a^2 = 0 may subtend a right angle at the center of the circle is:
A (a^{2} = 2p^{2})
B (p^{2} = 2a^{2})
C (a = 2p)
D (p = 2a)
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Answer
Question: The condition that the chord (x \cos \alpha + y \sin \alpha  p = 0) of (x^2 + y^2  a^2 = 0) may subtend a right angle at the center of the circle is:
 (A) (a^{2} = 2p^{2})
 (B) (p^{2} = 2a^{2})
 (C) (a = 2p)
 (D) (p = 2a)
****:
Given the circle equation (x^2 + y^2  a^2 = 0), we are asked about the chord (x \cos \alpha + y \sin \alpha  p = 0) and its condition to subtend a right angle at the circle's center.

Center and Radius: The center of the circle is at ( (0, 0) ) and the radius is ( a ).

Distance from Center to Chord: The perpendicular distance from the center ((0,0)) to the chord can be calculated using the formula:
$$ d = \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} $$
For the chord equation (x \cos \alpha + y \sin \alpha  p = 0), substituting (A = \cos \alpha), (B = \sin \alpha), and (C = p):
$$ d = \frac{\cos \alpha \cdot 0 + \sin \alpha \cdot 0  p}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = \frac{  p }{\sqrt{1}} = p $$

Subtending a Right Angle at the Center: For the chord to subtend a right angle at the center, denote the midpoint of chord (AB) as (M). We need to have ( \angle AMB = 90^\circ ).
Consider the rightangled triangle (AMC) where:
 (CA = a) (radius)
 (CM) is the perpendicular distance, i.e., ( p )
Since ( \angle AMB = 90^\circ ), the triangle is an isosceles right triangle, and therefore:
$$ CM = CA \cos 45^\circ $$
Given (\cos 45^\circ = \frac{1}{\sqrt{2}}):
$$ p = a \cos 45^\circ = \frac{a}{\sqrt{2}} $$

Squaring Both Sides:
$$ a^2 = 2p^2 $$
Hence, the correct condition is:
Answer: A) (a^{2} = 2p^{2})
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