Question

The limiting points of the coxial system determined by the circles
$x^2+y^2+4x-2y+5=0$ and$x^2+y^2+2x+4y+7=0$ are

• A. $(-2,-1),(0,-3)$
• B. $(-2,1),(0,-3)$
• C. $(-2,1),(0,3)$
• D. $(1,-2),(-3,0)$

Answer 1

The correct option is A $(-2,-1),(0,-3)$

The equation of circle of coaxial system is

$x^2+y^2+4x+2y+5+\lambda (x^2+y^2+2x+4y+7)=0$

$\Rightarrow x^2+y^2+\frac{4+2\lambda }{1+\lambda }x+\frac{2+4\lambda }{1+\lambda }y+\frac{5+7\lambda }{1+\lambda }=0$

The center and radius of circle are

$(-\frac{2+\lambda }{1+\lambda },-\frac{1+2\lambda }{1+\lambda })$ and $\sqrt{{\left(\frac{2+\lambda }{1+\lambda }\right)}^2+{\left(\frac{1+2\lambda }{1+\lambda }\right)}^2-\left(\frac{5+7\lambda }{1+\lambda }\right)}$

respectively

$\because$ Radius $=0$

$\Rightarrow \sqrt{\frac{-2{\lambda }^2-4\lambda }{{(1+\lambda )}^2}}=0\Rightarrow \lambda =0$ and $\lambda =-2$

$\lambda =0\Rightarrow$ center $(-2,-1)$;$\lambda =-2\Rightarrow$ center is $(0,-3)$