Question

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

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

Answer 1

The correct option is A 2 < r < 8

$\text{Circles intersect}if|r_1-r_2|<C_1{C}_2\Rightarrow |r-3|<5\Rightarrow 0<r<8...\left(i\right)and{r}_1+r_2>C_1{C}_2\Rightarrow r+3>5\Rightarrow r>2...\left(ii\right)$
From (i) and (ii), we get
2 < r < 8