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: [2025]

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Solution

Q.
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: [2025]

1 $2 \sqrt{2}$

2 $2 \sqrt{3}$

3 $4 \sqrt{2}$

4 $\sqrt{3}$

Ans.
(2)

Since, centre (h, k) lies on 3x + 2y + 2 = 0

$\Rightarrow$ 3h + 2k + 2 = 0 ... (i)

Also, the circle passes through the points (4, 2) and (0, 2), then we can say that ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ passes through (4, 2) and (0, 2)

$∴ h^{2} + {(2 - k)}^{2} = r^{2}$ ... (ii)

and ${(4 - h)}^{2} + {(2 - k)}^{2} = r^{2}$ ... (iii)

On subtracting (ii) from (iii), we get

$h^{2} = {(4 - h)}^{2} \Rightarrow h^{2} = 16 + h^{2} - 8 h \Rightarrow h = 2$

From (i), k = – 4

$∴$ Centre = (2, –4)

Radius, $r = \sqrt{{(0 - 2)}^{2} + {(2 + 4)}^{2}} = \sqrt{40}$

Now, mid-point of the chord is (1, 2)

$\Rightarrow$ Perpendicular distance from centre to chord = d

$= \sqrt{{(2 - 1)}^{2} + {(- 4 - 2)}^{2}} = \sqrt{37}$

$∴$ Length of chord = $2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{40 - 37} = 2 \sqrt{3}$

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