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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

\mathrm{2<\mathrm{r}<8}


Option: 2

r<2


Option: 3

r=2


Option: 4

r>2


Answers (1)

best_answer

Let distance between the centers of two circles of radii \mathrm{r_1 \text { and } r_{2^*}}

These circle intersect at two distinct points if  \mathrm{\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{d}<\mathrm{r}_1+\mathrm{r}_2}

Here, the radii of the two circles are r and 3 and distance between the centres is 5.

Thus,  \mathrm{|\mathrm{r}-3|<5 \mathrm{r}+3 \Rightarrow-2<\mathrm{r}<8 \text { and } \mathrm{r}>2 \Rightarrow 2<\mathrm{r}<8}

Hence (a) is correct answer. 

Posted by

manish painkra

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