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.

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

  2. 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):

    $$ d = \frac{|\cos \alpha \cdot 0 + \sin \alpha \cdot 0 - p|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = \frac{| - p |}{\sqrt{1}} = |p| $$

  3. 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 right-angled 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}}):

    $$ p = a \cos 45^\circ = \frac{a}{\sqrt{2}} $$

  4. Squaring Both Sides:

    $$ a^2 = 2p^2 $$

Hence, the correct condition is:

Answer: A) (a^{2} = 2p^{2})


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