Multiple choice

The line $x\cos\alpha+y\sin\alpha=p$ intersects the circle $x^{2}+y^{2}=4$ at $A$ and $B$. If the chord $AB$ makes an angle $30^{o}$ at a point on the circumference of the circle, then:

  1. $p^{2}=3$
  2. $p^{2}=4\cos^{4}15^{o}$
  3. $p^{2}=2$
  4. $p^{2}=6$
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A Correct answer
Explanation

The distance from the origin to the line x*cos(alpha) + y*sin(alpha) = p is p. For a circle x^2 + y^2 = r^2, the chord length is 2*sqrt(r^2 - p^2). The angle subtended by the chord at the circumference is 30 degrees, so the angle at the center is 60 degrees. This makes the triangle formed by the center and the chord an equilateral triangle with side length r=2. Thus, the distance p = r*cos(30) = 2*(sqrt(3)/2) = sqrt(3). So p^2 = 3.

AI explanation

The circle x^2 + y^2 = 4 has a radius of 2. A chord that subtends an angle of 30 degrees at a point on the circumference must subtend a central angle of 60 degrees. The perpendicular distance from the center to this chord is p = r * cos(central angle / 2) = 2 * cos(30 degrees) = 2 * (sqrt(3) / 2) = sqrt(3). Squaring this perpendicular distance gives p^2 = 3.

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