Question

If the product of the lengths of the perpendiculars drawn from the ends of a diameter of the circle \( x^2 + y^2 = 4 \) onto the line \( x + y + 1 = 0 \) is maximum, then the two ends of that diameter are:

• A. (−2, 0),(2, 0)
• B. (√3, 1),(−√3, −1)
• C. (√2,√2),(−√2, −√2)
• D. (0, 2),(0, −2)

Hint:

For optimization problems involving trigonometric functions, consider using trigonometric identities to simplify expressions.

Answer 1

The Correct Option is C

Step 1: Understand the Problem
We need to find the diameter of the circle \( x^2 + y^2 = 4 \) (radius = 2) such that the product of the perpendicular distances from its endpoints to the line \( x + y + 1 = 0 \) is maximized. Step 2: Parametrize the Diameter
Let the endpoints of the diameter be \( P(2\cos\theta, 2\sin\theta) \) and \( Q(-2\cos\theta, -2\sin\theta) \). Step 3: Compute Perpendicular Distances
The perpendicular distance from \( P \) to the line \( x + y + 1 = 0 \) is: \[ d_1 = \frac{|2\cos\theta + 2\sin\theta + 1|}{\sqrt{2}} \] Similarly, the distance from \( Q \) is: \[ d_2 = \frac{| -2\cos\theta - 2\sin\theta + 1 |}{\sqrt{2}} \] Step 4: Compute the Product \( d_1 \cdot d_2 \)
\[ d_1 \cdot d_2 = \frac{|1 - 4(\cos\theta + \sin\theta)^2|}{2} \] To maximize this, we need to maximize \( |1 - 4(\cos\theta + \sin\theta)^2| \). The maximum occurs when \( \cos\theta + \sin\theta = 0 \), giving the points \( (\sqrt{2}, \sqrt{2}) \) and \( (-\sqrt{2}, -\sqrt{2}) \). Conclusion:
The correct option is (3).