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

Answer: (B)

$3 - \sqrt{2}$

Answer 2

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: $3 - \sqrt{2}.$