Q210 / 270 · NDA 2019 (Phase 1) · Mathematics

The circle $x^2+y^2+4 x-7 y+12=0$, cuts an intercept on y -axis equal to

  • Correct answer
    A
    1
  • B
    3
  • C
    4
  • D
    7

Solution

Given circle, $x^2+y^2+4 x-7 y+12=0$ Comparing with general form of circle, $\begin{aligned} & \mathrm{ax}^2+\mathrm{by}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \\ & f=\frac{-7}{2} \text { and } \mathrm{c}=12\end{aligned}$ $\begin{aligned} & \mathrm{y}-\text { intercept }=2 \sqrt{f^2-c} \\ & =2 \sqrt{\left(\frac{-7}{2}\right)^2-12}=2 \sqrt{\frac{49}{4}-12} \\ & =2 \sqrt{\frac{49-48}{4}}=2\left(\frac{1}{2}\right)=1\end{aligned}$

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

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