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CONIC SECTIONS — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSCONIC SECTIONS1 Mark
MCQ
If $2\text{x}^2+\lambda\text{xy}+2\text{y}^2(\lambda-4)\text{x}+6\text{y}-5=0$ is the equation of a circle, then its radius is:
  • A
    $3\sqrt{2}$
  • B
    $2\sqrt{3}$
  • C
    $2\sqrt{2}$
  • ✓
    None of these
✓

Answer

Correct option: D.
None of these
The given equation is $2\text{x}^2+\lambda\text{xy}+2\text{y}^2+(\lambda-4)\text{x}+6\text{y}-5=0$ which can be rewritten as
$\text{x}^2+\frac{\lambda\text{xy}}{2}+\text{y}^2+\frac{(\lambda-4)}{2}\text{x}+3\text{y}-\frac{5}{2}=0.$
Comparing the given equation $\text{x}^2+\text{y}62+2\text{gx}+2\text{fy}+\text{c}=0$ with we get: $\lambda=0$
$\therefore\text{x}^2+\text{y}^2-2\text{x}+3\text{y}-\frac{5}{2}=0$
$\therefore$ Radius $=\sqrt{(-1)^2+\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$

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