If the 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

If the 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
  1. $r>2$
  2. $2 < \mathrm{r} < 8$
  3. $r < 2$
  4. $\mathrm{r}=2$

Solution

$\left|r_1-r_2\right|=C_1 C_2$ for intersection $\Rightarrow \mathrm{r}-3 < 5 \Rightarrow \mathrm{r} < 8$
and $\mathrm{r}_1+\mathrm{r}_2>\mathrm{C}_1 \mathrm{C}_2, \mathrm{r}+3>5 \Rightarrow \mathrm{r}=2$
From (1) and (2), $2 < \mathrm{r} < 8$.

Asked in: JEE Main 2003

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