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)

Key Idea : Two circles of radii $r_1$ and $r_2$ respectively intersect in two distinct points, if $r_1-r_2<C_1{C}_2<r_1+r_2$
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
$x^2+y^2-8x+2y+8=0$
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 $2<r$
$\Rightarrow 2<r<8$