1. Circle and chord PQ

The circle is centered at the origin with radius $\sqrt{10}$, so its equation is $x^2+y^2=10.$

The chord $PQ$ lies on the line $x+y=2.$ Since this line has slope $-1$, any chord parallel to it will also have equation of the form $x+y=c.$ So chord $MN$ must be of this form.

2. Use the length of chord formula

For the circle $x^2+y^2=10$, if a line is at perpendicular distance $d$ from the center, then the length of the chord cut by the line is $L=2\sqrt{r^2-d^2}.$ Here,

• $r=\sqrt{10}$
• chord length $L=2$

So, $2=2\sqrt{10-d^2}$ $\sqrt{10-d^2}=1$ $10-d^2=1$ $d^2=9$ $d=3.$

Thus, the line containing chord $MN$ is at distance $3$ from the origin.

3. Find equations of lines parallel to $x+y=2$ at distance 3 from origin

A line parallel to $x+y=2$ has equation $x+y=c.$ Its distance from the origin is $\frac{|c|}{\sqrt{1^2+1^2}}=\frac{|c|}{\sqrt{2}}.$

Set this equal to $3$: $\frac{|c|}{\sqrt{2}}=3$ $|c|=3\sqrt{2}.$ So possible lines are $x+y=3\sqrt{2} \quad \text{or} \quad x+y=-3\sqrt{2}.$

Either can represent chord $MN$.

4. Distance between chord PQ and chord MN

Chord $PQ$ lies on $x+y=2.$ A parallel chord $MN$ lies on either $x+y=3\sqrt{2} \quad \text{or} \quad x+y=-3\sqrt{2}.$

Distance between parallel lines $x+y=c_1$ and $x+y=c_2$ is $\frac{|c_1-c_2|}{\sqrt{2}}.$

If $MN$ is $x+y=3\sqrt{2}$, then distance from $PQ$ is $\frac{|3\sqrt{2}-2|}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}}-\frac{2}{\sqrt{2}}=3-\sqrt{2}.$

If $MN$ is $x+y=-3\sqrt{2}$, then distance is $\frac{| -3\sqrt{2}-2|}{\sqrt{2}}=3+\sqrt{2},$ which is not among the options.

Hence the required distance is $3-\sqrt{2}.$

5. Check options

• A: $3-\sqrt{2}$  correct
• B: $2-\sqrt{3}$  incorrect
• C: $\sqrt{2}-1$  incorrect
• D: $\sqrt{2}+1$  incorrect

Therefore, the correct option is A.