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

Option: 1

2<\mathrm{r}<8


Option: 2

\quad \mathrm{r}<2


Option: 3

\mathrm{r}=2


Option: 4

\mathrm{r}>2


Answers (1)

best_answer

Let \mathrm{d} be the distance between the centres of two circles of racdii \mathrm{r_{1}\: and \: r_{2}}.

These circle intersect attwo distinct points if  \mathrm{\left|r_{1}-r_{2}\right|<d<r_{1}+r_{2}}

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

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

Hence (a) is the correct answer.

Posted by

Kuldeep Maurya

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