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

$2\pm \frac{4\sqrt{2}}{3}$

The given circle is
$x^2+y^2+6x+5+\lambda (x^2+y^2-8x+7)=0or{x}^2(1+\lambda )+y^2(1+\lambda )+(6-8\lambda )x+(5+7\lambda )=0\Rightarrow x^2+y^2+\left(\frac{6-8\lambda }{1+\lambda }\right)x+\left(\frac{5+7\lambda }{1+\lambda }\right)=0$
This will dwindle into a point circle, then Radius of the circle = 0
$\sqrt{{\left(\frac{3-4\lambda }{1+\lambda }\right)}^2+0-\left(\frac{5+7\lambda }{1+\lambda }\right)=0}\Rightarrow (3-4\lambda {)}^2-(5+7\lambda \left)\right(1+\lambda )=0$
$\Rightarrow 9+16{\lambda }^2-24\lambda -5-5\lambda -7\lambda -7{\lambda }^2=0\Rightarrow 9{\lambda }^2-36\lambda +4=0\lambda =\frac{36\pm \sqrt{(36{)}^2-\mathrm{4.9.4}}}{2.9}\therefore \lambda =2\pm \frac{4\sqrt{2}}{3}$