If the circle $x^2+y^2+8 x-4 y+c=0$ touches the circle $x^2+y^2+2 x+4 y-11=0$ externally and cuts the circle…

If the circle $x^2+y^2+8 x-4 y+c=0$ touches the circle $x^2+y^2+2 x+4 y-11=0$ externally and cuts the circle $x^2+y^2-6 x+8 y+k=0$ orthogonally, then $k$ is equal to
  1. 59
  2. -59
  3. 19
  4. -19

Solution

Given that circle $x^2+y^2+8 x-4 y+c=0$ touch the circle $x^2+y^2+2 x+4 y-11=0$
where $\begin{gathered} C_1=(-4,2) \\ r_1=\sqrt{16+4-c}=\sqrt{20-c} \\ C_2=(-1,-2) \end{gathered}$ and $r_2=\sqrt{1+4+11}=4$ $\therefore$ From Eq. (i), $\begin{aligned} & \sqrt{(-4+1)^2+(2+2)^2} \\ & =\sqrt{20-c}+4 \\ & \Rightarrow \quad 5=\sqrt{20-c}+4 \\ & \Rightarrow \quad c=19 \\ & \end{aligned}$ Also, the circle $x^2+y^2+8 x-4 y+c=0$ cuts the circles $x^2+y^2-6 x+8 y+k=0$ orthogonally, then $c_1 \rightarrow(-4,2)$ $\begin{aligned} & C_3 \rightarrow(3,-4) \\ & \left(C_1 C_3\right)^2=\left(r_1\right)^2+\left(r_3\right)^2 \\ & \text { where, } \\ & r_1=\sqrt{16+4-c} \\ & r_3=\sqrt{9+16-k} \\ & \Rightarrow \\ & \left\{(-4-3)^2+(2+4)^2\right\}=(20-c)+(25-k) \\ & \Rightarrow \quad 49+36=45-k-c \\ & \Rightarrow \quad k+c=-40 \\ & \Rightarrow \quad k+19=-40 \\ & \Rightarrow \quad k=-59 \\ & \end{aligned}$

Asked in: AP EAMCET 2011

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