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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
Option: 1
$2<r<8$

Option: 2
$r<2$

Option: 3
$r=2$

Option: 4
$r>2$

Answers (1)

best_answer

Centres of the given circles are $C_1(1,3)$ and $C_2(4,-1)$ , and their radii, $r_1=r$ and $r_2=3.$

We know that the two circles touch,

externally if $\quad C_1 C_2=r_1+r_2$ , and internally if $\quad C_1 C_2=\left|r_1-r_2\right|$ .

Thus the two circles will cut at two distinct points if

$C_1 C_2>\left|r_1-r_2\right| \text { and } C_1 C_2<r_1+r_2$ $\text {, i.e., if }\left|r_1-r_2\right|<C_1 C_2<r_1+r_2 \text {, }$

or if $r-3<5<r+3$ or if $r<8$ and $r>2$ or if , which is given in (A).

$\therefore \quad (a)$

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