Let C be a circle with radius √10 units and centre at the origin. Let the line x + y = 2 intersects

Q: Wait, why do we have two possible answers ($d_1 + d_2$ and $|d_1 - d_2|$)? How do I know which one to pick?

That is a fantastic question! When you have two parallel chords in a circle, they can either lie on the same side of the center or on opposite sides. If they are on the same side, the distance between them is the difference of their distances from the center, $|d_1 - d_2|$. If they are on opposite sides, the distance is the sum, $d_1 + d_2$. In this problem, we look at the provided options to see which one fits, which confirms they are on the same side.

Q: Why is the slope of the chord $MN$ equal to the slope of $PQ$?

The problem explicitly states that the chord $MN$ has a slope of $-1$. If you rearrange the equation of the line $PQ$, which is $x + y = 2$, into the slope-intercept form $y = -x + 2$, you can clearly see its slope is also $-1$. Since both chords have the same slope, they are geometrically parallel.

Q: Can I just find the equation of the line $MN$ instead of using the distance formula?

You certainly could, but it would be much harder! To find the equation of $MN$, you would need to find the specific point on the circle where a chord of length $2$ exists with slope $-1$. Using the distance formula is a much more elegant and direct path to the solution, as it bypasses the need to find the coordinates of $M$ and $N$ entirely.

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

Options:

(A) $3 - \sqrt{2}$

(B) $\sqrt{2} + 1$

(C) $\sqrt{2} - 1$

(D) $2 - \sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the origin of a coordinate plane, looking out at a circle $C$ with a radius of $R = \sqrt{10}$. This circle is our universe for this problem.

We are given two chords, $PQ$ and $MN$, and we need to find the distance between them. Let us break this down into a logical sequence.

### Phase 1: The First Chord ($PQ$)

Our first chord, $PQ$, is defined by the line $x + y = 2$. To understand its position, we calculate its perpendicular distance from the center $(0,0)$.

We use the perpendicular distance formula:
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Here, our line is $x + y - 2 = 0$, so $A=1$, $B=1$, and $C=-2$. Plugging in the origin $(0,0)$:
$$d_1 = \frac{|1(0) + 1(0) - 2|}{\sqrt{1^2 + 1^2}} = \frac{|-2|}{\sqrt{2}} = \sqrt{2}$$

Thus, our first chord $PQ$ is exactly $\sqrt{2}$ units away from the center.

### Phase 2: The Second Chord ($MN$)

Now, let us turn our attention to the second chord, $MN$. We are told its length is $2$ units and its slope is $-1$.

Since the slope of $PQ$ is also $-1$, the chords $MN$ and $PQ$ are parallel. To find the distance $d_2$ of $MN$ from the center, we use the property that a perpendicular from the center bisects the chord.

This creates a right-angled triangle where the hypotenuse is the radius $R = \sqrt{10}$, one leg is $d_2$, and the other leg is half the chord length, which is $\frac{2}{2} = 1$. By the Pythagorean theorem:
$$R^2 = d_2^2 + \left(\frac{L}{2}\right)^2$$

Substituting our values:
$$(\sqrt{10})^2 = d_2^2 + 1^2$$
$$10 = d_2^2 + 1$$
$$d_2^2 = 9 \Rightarrow d_2 = 3$$

### Phase 3: The Final Synthesis

We now have two parallel chords. One is at a distance of $d_1 = \sqrt{2}$ from the center, and the other is at a distance of $d_2 = 3$ from the center.

The distance between these two parallel chords is either the difference of their distances (if they are on the same side of the center) or the sum of their distances (if they are on opposite sides). This gives us two possibilities: $3 - \sqrt{2}$ or $3 + \sqrt{2}$.

The final distance between the chords is **$3 - \sqrt{2}$**. This result confirms that both chords lie on the same side of the center.

Select Answer:

Visualized Solution

Visualizing the Circle $C$

Analyzing Chord $P Q$

Distance $d_{1}$ Setup

Calculating $d_{1}$

Properties of Chord $M N$

Distance $d_{2}$ from Center to $M N$

Setting up Pythagoras for $M N$

Calculating $d_{2}$

Distance Between Parallel Chords

Matching with Options

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Visualizing the Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Analyzing Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​ Setup

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​

Properties of Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN

Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​ from Center to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN

Setting up Pythagoras for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } MN

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2​