Let a circle C pass through the points (4,2) and (0, 2), and its centre lie on y+2=0 Then the length

Q: Wait, why do we reject the negative root here?

Actually, we don't reject roots; we are dealing with lengths. The distance formula gives us the square of the distance, which is always positive. When we take the square root, we only consider the positive value because a length cannot be negative.

Q: Why can't we just use the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$?

You absolutely can! It is a valid path. However, using the geometric properties of the perpendicular bisector is often faster and less prone to algebraic errors. In JEE, speed and accuracy are your best friends.

Q: Is the midpoint of the chord always connected to the center by a perpendicular line?

Yes, this is a fundamental theorem in circle geometry. The line segment joining the center of a circle to the midpoint of a chord is always perpendicular to the chord. This property is the key to unlocking many chord-related problems.

Question

Let a circle C pass through the points $(4,2)$ and (0, 2), and its centre lie on $y+2=0$ Then the length of the chord, of the circle C, whose mid-point is $(1,2)$, is:

Options:

(A) $\sqrt{3}$

(B) $2\sqrt{2}$

(C) $2\sqrt{3}$

(D) $4\sqrt{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two points: $A(4, 2)$ and $B(0, 2)$. They both sit on the horizontal line $y=2$.

Because $A$ and $B$ form a horizontal chord, their perpendicular bisector must be a vertical line passing through their midpoint. The midpoint of $AB$ is calculated as:

$$M_{AB} = \left( \frac{4+0}{2}, \frac{2+2}{2} ight) = (2, 2)$$

Thus, the perpendicular bisector is the vertical line $x=2$. Since the center of any circle must lie on the perpendicular bisector of any of its chords, the center $C$ must lie on the line $x=2$.

### Finding the Heart of the Circle

The problem states that the center lies on the line $y+2=0$, which is $y=-2$. We now have a classic intersection problem.

The intersection of the lines $x=2$ and $y=-2$ gives us the center of the circle: $C(2, -2)$.

With the center $C(2, -2)$ and a point on the circle $B(0, 2)$, we calculate the radius squared, $r^2$, using the distance formula:

$$r^2 = (2-0)^2 + (-2-2)^2 = 2^2 + (-4)^2 = 4 + 16 = 20$$

This value, $r^2 = 20$, serves as the anchor for our entire calculation.

### The Target Chord: A Pythagorean Journey

We are now tasked with finding the length of a chord, $L$, whose midpoint is $M(1, 2)$. The secret to any chord problem is the right-angled triangle formed by the center, the midpoint of the chord, and one of the chord's endpoints.

Let $P$ be an endpoint of the chord. In the right-angled triangle $	riangle CMP$, $CM$ is the perpendicular distance from the center to the chord, $MP$ is half the length of the chord ($L/2$), and $CP$ is the radius $r$.

First, we calculate the distance $d = CM$ using $C(2, -2)$ and $M(1, 2)$:

$$d^2 = (2-1)^2 + (-2-2)^2 = 1^2 + (-4)^2 = 1 + 16 = 17$$

Next, we apply the Pythagorean theorem:

$$r^2 = d^2 + \left( \frac{L}{2} ight)^2$$

Substituting our known values, we get:

$$20 = 17 + \left( \frac{L}{2} ight)^2 \Rightarrow \left( \frac{L}{2} ight)^2 = 3$$

Taking the square root, we find the half-length of the chord is $\sqrt{3}$. Finally, the total length $L$ is:

**$L = 2\sqrt{3}$**

Animated Solution for Mathematics - Circles: Let a circle C pass through the points $(4, 2)$ and (0, 2), and its centre lie on $y + 2 = 0$ Then the length of the chord, of the circle C, whose mid-point is $(1, 2)$ , is:

Select Answer:

Visualized Solution

Identify Given Points

Perpendicular Bisector of $A B$

Find the Center of the Circle

Calculate Radius Squared ( $r^{2}$ )

Identify the Target Chord

Distance from Center to Midpoint

Apply Pythagoras Theorem

Calculate Half-Length of Chord

Final Chord Length

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Finding the Heart of the Circle

The Target Chord

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