Q64 / 180 · KCET 2008 · Mathematics

The equation of the smallest circle passing through the points $(2,2)$ and $(3,3)$ is

  • A
    $x^{2}+y^{2}+5 x+5 y+12=0$
  • Correct answer
    B
    $x^{2}+y^{2}-5 x-5 y+12=0$
  • C
    $x^{2}+y^{2}+5 x-5 y+12=0$
  • D
    $x^{2}+y^{2}-5 x+5 y-12=0$

Solution

Let the points be $A=(2,2)$ and $B=(3,3)$. Since, the circle passing through these points, so they satisfie the equation of the circle. Now, taking option (b), Let $\mathrm{S} \equiv \mathrm{x}^{2}+\mathrm{y}^{2}-5 \mathrm{x}-5 \mathrm{y}+12=0$ At $\mathrm{A}=(2,2)$ $2^{2}+2^{2}-5(2)-5(2)+12=0$ $\Rightarrow$ $0=0$ At $\mathrm{B}=(3,3)$ $3^{2}+3^{2}-5(3)-5(3)+12=0$ $\Rightarrow$ $0=0$ Hence, option (b) is correct.

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

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