Q190 / 270 · NDA 2016 (Phase 1) · Mathematics

Consider the two circles $(\mathrm{x}-1)^2+(\mathrm{y}-3)^2=\mathrm{r}^2$ and $x^2+y^2-8 x+2 y+8=0$ What is the distance between the centres of the two circles?

  • Correct answer
    A
    5 units
  • B
    6 units
  • C
    8 units
  • D
    10 units

Solution

Given equation of circles $\begin{aligned} & (x-1)^2+(y-3)^2=r^2 \\ & \left(h_1, k_1\right) \equiv \text { coordinates of centre } \equiv(1,3) \\ & \therefore x^2+y^2-8 x+2 y+8=0 \\ & \Rightarrow(x-4)^2+(y+1)^2=(3)^2 \\ & \left(h_2, k_2\right) \equiv \text { coordinates of centre } \equiv(4,-1) \end{aligned}$ Distance between centres of two circles $\begin{aligned} & d=\sqrt{\left(h_1-h_2\right)^2+\left(k_1-k_2\right)^2} \\ & \Rightarrow d=\sqrt{(1-4)^2+(3+1)^2} \\ & \Rightarrow d=\sqrt{25} \\ & d=5 \text { units } \end{aligned}$

Hence, the correct answer is option: 'A'.

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