JEE MAIN - Mathematics (2025 - 29th January Evening Shift - No. 12)
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:
4$\sqrt{2}$
2$\sqrt{2}$
2$\sqrt{3}$
$\sqrt{3}$
Explanation
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$$\begin{aligned} &\begin{aligned} & \mathrm{M}_{A B}=0 \Rightarrow \mathrm{OM} \text { is vertical } \\ & \Rightarrow \alpha=2 \\ & \therefore \text { Centre }(0) \equiv(2,-4) \\ & \quad r=O A=\sqrt{(2-4)^2+(2+4)^2}=\sqrt{40} \end{aligned}\\ &\text { mid point of chord is } \mathrm{N} \equiv(1,2) \quad \therefore \mathrm{ON}=\sqrt{37}\\ &\begin{aligned} \therefore \text { length of chord } & =2 \sqrt{\mathrm{r}^2-(\mathrm{ON})^2} \\ & =2 \sqrt{40-37}=2 \sqrt{3} \end{aligned} \end{aligned}$$
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