JEE MAIN - Mathematics (2024 - 5th April Evening Shift - No. 3)
Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :
$$\mathrm{r}=1$$
$$2 \mathrm{r}^2-4 \mathrm{r}+1=0$$
$$2 \mathrm{r}^2-8 \mathrm{r}+7=0$$
$$\mathrm{r}^2-8 \mathrm{r}+8=0$$
Explanation
_5th_April_Evening_Shift_en_3_1.png)
$$C F=4 \sqrt{2}-2 \sqrt{2}=2 \sqrt{2}=r+r \sqrt{2}$$
$$\begin{aligned} & \Rightarrow \quad(2-r) \sqrt{2}=r \\ & \Rightarrow \quad \sqrt{2}=\left(\frac{r}{2-r}\right) \Rightarrow 2=\frac{r^2}{(2-r)^2} \\ & \Rightarrow 2\left(r^2-4 r+4\right)=r^2 \\ & \Rightarrow r^2-8 r+8=0 \end{aligned}$$
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