If 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, the

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10-1.Circle and System of Circles

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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

A

$5<$ r $<9$

B

$0<$ r $<7$

C

$3<$ r $<7$

D

$\frac{1}{2}<$ r $<7$

(JEE MAIN-2024)

Solution

If two circles intersect at two distinct points

$\Rightarrow\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2$

$ |\mathrm{r}-2|<\sqrt{9+16}<\mathrm{r}+2 $

$ |\mathrm{r}-2|<5 \text { and } \mathrm{r}+2>5$

$ -5<\mathrm{r}-2<5$

$r>3$ $………….(1)$

$-3<\mathrm{r}<7$ $………………(2)$

From $(1)$ and $(2)$

$3<\mathrm{r}<7$

Standard 11

Mathematics

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