Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is:

Circle PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q37 with Solution

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Correct answer: C

The circle passes through $(4,2)$ and $(0,2)$ .
So the centre must lie on the perpendicular bisector of the chord joining these points.

The midpoint of $(4,2)$ and $(0,2)$ is

\left(\frac{4+0}{2},\frac{2+2}{2}\right)=(2,2).

Since the segment joining the points is horizontal, its perpendicular bisector is the vertical line

x=2.

Also, the centre lies on

3x+2y+2=0.

Substitute $x=2$ :

3(2)+2y+2=0

6+2y+2=0

2y+8=0

y=-4.

Hence the centre is

O=(2,-4).

Radius is the distance from the centre to either given point, say $(4,2)$ :

r=\sqrt{(4-2)^2+(2+4)^2} =\sqrt{2^2+6^2} =\sqrt{40}=2\sqrt{10}.

So,

r^2=40.

In a circle, the line joining the centre to the midpoint of a chord is perpendicular to the chord.

Given midpoint of the required chord is

M=(1,2).

Distance from centre to this midpoint is

OM=\sqrt{(1-2)^2+(2+4)^2} =\sqrt{(-1)^2+6^2} =\sqrt{37}.

If the distance from centre to chord is $d$ , then chord length is

2\sqrt{r^2-d^2}.

Here,

d=\sqrt{37}, \quad r^2=40.

Therefore,

\text{Chord length}=2\sqrt{40-37}=2\sqrt{3}.

2\sqrt{3}

corresponds to Option C.

Final Answer: $\boxed{2\sqrt{3}}$