Question

P and Q are the ends of a diameter of the circle \( x^2+y^2=a^2(a>\frac{1}{\sqrt{2}}) \). \( s \) and \( t \) are the lengths of the perpendiculars drawn from P and Q onto the line \( x+y=1 \) respectively. When the product \( st \) is maximum, the greater value among \( s, t \) is:

• A. \( a+\sqrt{2} \)
• B. \( a+\frac{1}{\sqrt{2}} \)
• C. \( a-\frac{1}{\sqrt{2}} \)
• D. \( a-\sqrt{2} \)

Hint:

For distance problems involving perpendiculars, use: \[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] to compute distances effectively.

Answer 1

The Correct Option is B

The equation of the circle is: \[ x^2 + y^2 = a^2 \] The ends of the diameter satisfy \( (x_1, y_1) \) and \( (x_2, y_2) \), where: \[ (x_1, y_1) = (-a, a), (x_2, y_2) = (a, -a) \] The perpendicular distances of these points from the line \( x + y = 1 \) are given by: \[ s = \frac{|(-a) + a - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \] \[ t = \frac{|a - a - 1|}{\sqrt{1^2 + 1^2}} = \frac{|-1|}{\sqrt{2}} = \frac{1}{\sqrt{2}} \] When \( st \) is maximized, we compute: \[ s + t = a + \frac{1}{\sqrt{2}} \] Thus, the correct answer is: \[ a + \frac{1}{\sqrt{2}} \]