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

Answer 1

The correct option is D 2 < r < 8

Circle intersect at two district points -

$\Rightarrow c_1{c}_2<r_1+r_2$

$\Rightarrow \sqrt{3^2+4^2}<r+3$

$\Rightarrow 5<r+3$

$\Rightarrow r>2$

Also, $r_1-r_2<c_1{c}_2$

$\Rightarrow r-3<5$

$\Rightarrow r<8$

$\therefore 2<r<8$

Hence, the answer is $2<r<8.$