Q97 / 150 · MHT CET 2020 (16 Oct Shift 1) · Mathematics

The radius of the circle passing through the points $(5,7),(2,-2)$ and $(-2,0)$ is

  • A
    2 units
  • Correct answer
    B
    5 units
  • C
    4 units
  • D
    3 Units

Solution

(B) Let $(\mathrm{h}, \mathrm{k})$ be the center of the circle which passes through $(5,7),(2,-2)$ and $(-2,0)$ $\therefore(h-2)^{2}+(k+2)^{2}=(h+2)^{2}+k^{2}$ $\therefore-4 \mathrm{~h}+4+4 \mathrm{k}+4=4 \mathrm{~h}+4 \quad \Rightarrow 8 \mathrm{~h}-4 \mathrm{k}=4 \quad \Rightarrow 2 \mathrm{~h}-\mathrm{k}=1$...(1) Also $(\mathrm{h}-5)^{2}+(\mathrm{k}-7)^{2}=(\mathrm{h}+2)^{2}+\mathrm{k}^{2}$ $\therefore-10 \mathrm{~h}+25-14 \mathrm{k}+49=4 \mathrm{~h}+4 \Rightarrow 14 \mathrm{~h}+14 \mathrm{k}=70 \Rightarrow \mathrm{h}+\mathrm{k}=5$...(2) Solving $(1)$ and $(2)$, we get, $\mathrm{h}=2, \mathrm{k}=3 \Rightarrow$ centre $=(2,3)$ $\therefore$ Radius $=3-(-2)=5$

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

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