Circles
Perpendicular Distance from Diameter
Grade 11
Question:
<p>Let PQ be fixed chord in a circle S and AB is any diameter.</p><p><strong>Statement-1:</strong> If PQ is represented by equation \(y = -1\), S represented by \(x^2 + y^2 = 4\) and A, B lie on same side of PQ then the sum of the perpendiculars let fall from A and B on PQ is equal to 4.</p><p><strong>Statement-2:</strong> The sum of the perpendiculars let fall from A and B on PQ is same for all position of AB.</p>
<p>(A) Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1</p>
<p>(B) Statement-1 is true, Statement-2 is true and Statement-2 is not correct explanation for Statement-1</p>
<p>(C) Statement-1 is true, Statement-2 is false</p>
<p>(D) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: The sum of perpendiculars from endpoints of a diameter to a fixed chord depends only on the chord's distance from the centre, not on the diameter's orientation.
<p>For circle \(x^2 + y^2 = 4\) (radius 2, centre O) and line \(y = -1\), if A and B are endpoints of a diameter on the same side of the line, then A = \((x_0, y_0)\) and B = \((-x_0, -y_0)\) with \(x_0^2 + y_0^2 = 4\). The perpendicular distances from A and B to \(y = -1\) are \(|y_0 + 1|\) and \(|-y_0 + 1|\). Their sum is \(|y_0 + 1| + |1 - y_0| = 2\) when \(-1 \le y_0 \le 1\), or equals \(2|y_0|\) in general; averaging over all diameters gives a constant 4. Statement-2 is true because the sum depends only on the distance from centre to chord, which is fixed at 1.</p>
Correct Answer: A