Question

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

• A. $2<r<8$
• B. $r<2$
• C. $r=2$
• D. $r>2$

Answer 1

Answer: (A)

As, the two circles intersect in two distinct points
$\Rightarrow$ Distance between centres lies between $|r_1-r_2|$ and $|r_1+r_2|.$
i.e. $|r-3|<\sqrt{(4-1{)}^2+(-1-3{)}^2}<|r+3|$
$\Rightarrow |r-3|<5<|r+3|$
$\Rightarrow r<8$ or $r>2$
$\therefore 2<r<8$