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Q.
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 points, then :
AIEEEAIEEE 2003Conic Sections
Solution:
Key Idea : Two circles of radii $r_{1}$ and $r_{2}$ respectively intersect in two distinct points, if $r_{1}-r_{2}
The equation of first circle is
$(x-1)^{2}+(y-3)^{2}=r^{2}$
Centre $C_{1}(1,3)$ and radius $r_{1}=r$
and equation of second circle is
$x^{2}+y^{2}-8 x+2 y+8=0$
Centre $C_{2}(4,-1)$ and radius $r_{2}=\sqrt{4^{2}+1^{2}-8}$
$=\sqrt{17-8}=3$
Two circles intersect in two distinct points, then
$ r_{1}-r_{2} < C_{1} C_{2} < r_{1}+r_{2} $
$\Rightarrow r-3 < \sqrt{(4-1)^{2}+(-1-3)^{2}} < r+3 $
$\Rightarrow r-3 < \sqrt{9+16} < r+3 $
$\Rightarrow r-3 < 5 < r+3 $
$\Rightarrow r-3 < 5$ and $ 5 < r + 3 $
$\Rightarrow r < 8 $ and $ 2 < r $
$\Rightarrow 2 < r < 8$