Question

The members of a family of circles are given by the equation $2(x^2+y^2)+\lambda x-(1+{\lambda }^2)y-10=0$. The number of circle(s) from this family that is/are cut orthogonally by the circle $x^2+y^2+4x+6y+3=0$ is

• A. $2$
• B. $1$
• C. $0$
• D. infinite

Answer 1

The correct option is A $2$

$(x^2+y^2)+\frac{\lambda }{2}x-\frac{(1+{\lambda }^2)}{2}y-5=0x^2+y^2+4x+6y+3=0$
They cut orthogonally,
$2g_1{g}_2+2f_1{f}_2=c_1+c_2\Rightarrow 2\cdot \frac{\lambda }{4}\cdot 2+2\cdot (-\frac{(1+{\lambda }^2)}{4})\cdot 3=-5+3\Rightarrow 2\lambda -3(1+{\lambda }^2)=-4\Rightarrow 3{\lambda }^2-2\lambda -1=0\Rightarrow 3{\lambda }^2-3\lambda +\lambda -1=0\Rightarrow (\lambda -1)(3\lambda +1)=0$
Two real roots