Q243 / 270 · NDA 2013 (Phase 2) · Mathematics

The radius of the circle $x^2+y^2+x+c=0$ passing through the origin is

  • A
    $\frac{1}{4}$
  • Correct answer
    B
    $\frac{1}{2}$
  • C
    1
  • D
    2

Solution

$x^2+y^2+x+c=0$ As, this circle is passing through origin, so $0+0+0+c=0 \Rightarrow c=0$ Hence, equation reduced to $x^2+y^2+x=0$ Radius of the circle $\mathrm{R}=\sqrt{g^2+f^2-c}$ Here, $2 g=1 \Rightarrow g=\frac{1}{2}, f=0, c=0$ Radius $\mathrm{R}=\sqrt{\left(\frac{1}{2}\right)^2+0^2+0^2}=\frac{1}{2}$.

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

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