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 $\frac {23}{4} * \pi$

  • $\begingroup$ 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

Complete squares after putting $\;\lambda xy=0\implies \lambda =0\;$:



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



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.