Question

The values of $\lambda$ for which the circle $x^2+y^2+6x+5+\lambda (x^2+y^2-8x+7)=0$ dwindles into a point are

• A. $1\pm \frac{\sqrt{2}}{3}$
• B. $2\pm \frac{2\sqrt{2}}{3}$
• C. $2\pm \frac{4\sqrt{2}}{3}$
• D. $1\pm \frac{4\sqrt{2}}{3}$

Answer 1

The correct option is C $2\pm \frac{4\sqrt{2}}{3}$

$x^2+y^2+6x+5+\lambda (x^2+y^2-8x+7)=0$ dwindles into a point when radius is zero
$\Rightarrow x^2+y^2+\left(\frac{3-4\lambda }{1+\lambda }\right)2x+\frac{5+7\lambda }{1+\lambda }=0$
Now, radius $\sqrt{g^2+f^2-c}=\sqrt{{\left(\frac{3-4\lambda }{1+\lambda }\right)}^2-\frac{5+7\lambda }{1+\lambda }}=0$
$\Rightarrow 9+16{\lambda }^2-24\lambda -5-7\lambda -5\lambda -7{\lambda }^2=0$
$\Rightarrow 9{\lambda }^2-36\lambda +4=0$
Therefore, $\lambda =2\pm \frac{4\sqrt{2}}{3}$
Ans: C