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Q. Equation of circle- $2x^2+ \lambda xy+2y^2+( \lambda -4)x+6y-5=0$ find area of the circle.

Attempt- For converting the equation from second degree to first degree $\lambda xy=0$.

Thus, $\lambda =0$ and-

$$(\lambda -4)x = 2gx$$ $$ 6y=2fy$$ $$c=-5$$ $$g=-2, f=3, c=-5$$

Radius of circle = $\sqrt{4+9+5}=\sqrt{18}$

Area of circle= $ \pi *18$

But the answer is $\frac {23}{4} * \pi$

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  • $\begingroup$ Your argument would be correct if the coefficients of $x^2$ and $y^2$ were $1$. But they are not. $\endgroup$ – Leo163 Nov 20 '16 at 11:56
  • $\begingroup$ Do they need to be equal? Correct me but shouldn't just their coefficient be equal? To satisfy $a=b$ where a and b are coefficients of x and y respectively $\endgroup$ – Akshat Batra Nov 20 '16 at 11:58
  • $\begingroup$ @AkshatBatra Yes, the coefficients of the quadratic expressions for $\;x\,,\,\,y\;$ must be equal if we have a circle (otherwise it is an ellipse), but then you must divide through the whole equation by that common coefficient, and that affects the radius...! $\endgroup$ – DonAntonio Nov 20 '16 at 11:59
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Complete squares after putting $\;\lambda xy=0\implies \lambda =0\;$:

$$0=2x^2+2y^2-4x+6y-5=2(x-1)^2-2+2\left(y-\frac32\right)^2-\frac92-5\implies$$

$$\implies2(x-1)^2+2\left(y-\frac32\right)^2=\frac{23}2\implies(x-1)^2+\left(y-\frac32\right)^2=\frac{23}4$$

and we have a circle of radius $\;\sqrt{\frac{23}4}\;$ , so its area is

$$\pi\sqrt{\frac{23}4}^2=\frac{23\pi}4$$

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