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 equation of the circle is:
\(x^2 + y^2 = 10\)

The line \(x + y = 2\) intersects the circle. Its perpendicular distance from the center \((0, 0)\) is calculated as:

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

Let \(MN\) be another chord with length 2 units and slope \(-1\). For the chord \(MN\), the midpoint divides it symmetrically, with length 2. Using geometry: 
\(MN = 2 \implies\) Half-length: \(AN = \frac{MN}{2} = 1.\)

In \(\triangle OAN\), using the Pythagoras theorem:

\[ ON^2 = OA^2 + AN^2 \quad \text{where} \quad OA = 3. \] \[ 10 = (OA)^2 + 1^2 \implies OA = 3. \]

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

Distance from center to \(PQ\): \[ \frac{|0 + 0 - 2|}{\sqrt{2}} = \sqrt{2}. \]

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

The perpendicular distance is the sum: \[ d = OA + \sqrt{2} = 3 + \sqrt{2}. \] Thus, the final distance between the two chords is: \[ 3 - \sqrt{2}. \]