JEE MAIN - Mathematics (2024 - 8th April Evening Shift - No. 19)
If the image of the point $$(-4,5)$$ in the line $$x+2 y=2$$ lies on the circle $$(x+4)^2+(y-3)^2=r^2$$, then $$r$$ is equal to:
2
3
4
1
Explanation
$$\begin{aligned} & \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \\ & \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \\ & \therefore \quad \text { Image is }\left(\frac{-28}{5}, \frac{9}{5}\right) \end{aligned}$$
Image lies on circle $$(x+4)^2+(y-3)^2=r^2$$
$$\begin{aligned} & \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \\ & \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \\ & \Rightarrow r=2 \end{aligned}$$
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