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

If the two circles $(x-1)^2+(y-3)^2=r^2$ and $x^2+y^2-8 x+2 y+8=0$ intersect at two distinct points, then
  1. 2 < r < 8
  2. $1 < r < 9$
  3. r = 2
  4. r = 8

Solution

Equation of given circles $ \begin{aligned} S_1:(x-1)^2+(y-3)^2 & =r^2 \\ \text { and } & S_2:(x-4)^2+(y+1)^2=3^2 \end{aligned} $ If circles $S_1$ and $S_2$ intersect at two distinct points, then $ \left|r_1-r_2\right| < C_1 C_2 < \left(r_1+r_2\right) $ $ \begin{aligned} & \text { So, }|r-3| < \sqrt{(4-1)^2+(-1-3)^2} < r+3 \\ & \Rightarrow(r-3) < 5 < r+3 \\ & \Rightarrow r+3>5 \text { and }|r-3| < 5 \\ & \Rightarrow r>2 \text { and } r-3 \in(-5,5) \\ & \Rightarrow r \in(-2,8) \\ & \Rightarrow 2 < r < 8 \end{aligned} $ Hence, option (a) is correct

Asked in: AP EAMCET 2019 (20 Apr Shift 2)

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