JEE Main 2025 Jan 29 Shift 2, Mathematics Q14: Circle: Chord, Tangent And Normal

JEE Main2025Jan 29, Shift 2Mathematics

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

A
$3$
B
$23$
C
$42$
D
$22$

Solution

Both given points $(4, 2)$ and $(0, 2)$ share the same $y$ -coordinate, so the perpendicular bisector of the chord joining them is the vertical line $x = 2$ . The centre lies on this bisector and on $3 x + 2 y + 2 = 0$ :

$x = 2 \Rightarrow 6 + 2 y + 2 = 0 \Rightarrow y = - 4$ .

Centre $O = (2, - 4)$ , radius $r = (4 - 2)^{2} + (2 - (- 4))^{2} = 4 + 36 = 40$ .

Distance from $O$ to the midpoint $N = (1, 2)$ : $O N = 1 + 36 = 37$ .

Length of chord $= 2 r^{2} - O N^{2} = 2 40 - 37 = 23$ .

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JEE Main 2025 Jan 29 Shift 2, Mathematics Q14: Circle: Chord, Tangent And Normal

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