Question

Let C be a circle with radius \( \sqrt{10} \) units and centre at the origin. Let the line \( x + y = 2 \) intersects the circle C at the points P and Q. Let MN be a chord of C of length 2 unit and slope \(-1\). Then, a distance (in units) between the chord PQ and the chord MN is

• A. \( 2 - \sqrt{3} \)
• B. \( 3 - \sqrt{2} \)
• C. \( \sqrt{2} - 1 \)
• D. \( \sqrt{2} + 1 \)

Answer 1

The Correct Option is B

The equation of the circle is:

The circle's equation is: \(x^2 + y^2 = 10\). The line \(x + y = 2\) intersects the circle. The perpendicular distance from the center \((0, 0)\) to this line is:

Distance from center to line: \[ \frac{|0 + 0 - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \]

Consider another chord \(MN\) with length 2 units and slope \(-1\). The midpoint of \(MN\) divides it symmetrically. Given \(MN = 2\), the half-length is \(AN = \frac{MN}{2} = 1\). In \(\triangle OAN\), by the Pythagorean theorem, \(ON^2 = OA^2 + AN^2\). With \(ON^2 = 10\) and \(AN = 1\), we get \(10 = OA^2 + 1^2\), which yields \(OA = 3\). This \(OA\) represents the distance from the center to the chord \(MN\).

Step 1: Perpendicular distance from the center to chord \(PQ\):

The distance from the center to \(PQ\) is \(\sqrt{2}\).

Step 2: Perpendicular distance between \(MN\) and \(PQ\):

The perpendicular distance between the chords is the difference between their distances from the center. Since \(MN\) has a distance of 3 from the center and \(PQ\) has a distance of \(\sqrt{2}\) from the center, the distance between the chords is \(3 - \sqrt{2}\).