JEE MAIN - Mathematics (2024 - 30th January Morning Shift - No. 3)
If the circles $$(x+1)^2+(y+2)^2=r^2$$ and $$x^2+y^2-4 x-4 y+4=0$$ intersect at exactly two distinct points, then
$$\frac{1}{2}<\mathrm{r}<7$$
$$3<\mathrm{r}<7$$
$$5<\mathrm{r}<9$$
$$0<\mathrm{r}<7$$
Explanation
If two circles intersect at two distinct points
$$\begin{aligned} & \Rightarrow\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<\sqrt{9+16}<\mathrm{r}+2 \\ & |\mathrm{r}-2|<5 \text { and } \mathrm{r}+2>5 \\ & -5<\mathrm{r}-2<5 \quad \mathrm{r}>3 ~\text{......... 2} \end{aligned}$$
$$-3<\mathrm{r}<7\quad$$ .... (1)
From (1) and (2)
$$3<\text { r }<7$$
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