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If two circles \mathrm{(x-1)^2+(y-3)^2=r^2}  and \mathrm{x^2+y^2-8 x+2 y+8=0}  intersect in two distinct points, then
 

Option: 1

2<\mathrm{r}<8


Option: 2

\mathrm{r<2}


Option: 3

\mathrm{r=2}


Option: 4

\mathrm{r>2}


Answers (1)

best_answer

Centres and radii of the given circles are \mathrm{C_1(1,3), r_1=r}  and \mathrm{C_2=(4,-1), r_2=3}  respectively since circles intersect in two distinct points, then

\begin{aligned} & \left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{C}_1 \mathrm{C}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & \Rightarrow|\mathrm{r}-3|<5<\mathrm{r}+3 \quad \quad \quad \quad \dots(i) \end{aligned}
from last two relations, \mathrm{r>2}
from firs two relations

\begin{aligned} & |\mathrm{r}-3|<5 \Rightarrow-5<\mathrm{r}-3<5 \\ & \Rightarrow-2<\mathrm{r}<8 \quad \quad \quad \quad \dots (ii) \end{aligned}
from eqs. (i) and (ii), we get 2<\mathrm{r}<8

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manish

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