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 circles belonging to the family that are cut orthogonally by the fixed circle $x^2+y^2+4x+6y+3=0$ is

• A. $2$
• B. $1$
• C. $0$
• D. none of these

Answer 1

The correct option is A $2$

$C_1=(\frac{-\lambda }{2\times 2},\frac{1+{\lambda }^2}{2\times 2})=(\frac{-\lambda }{4},\frac{{\lambda }^2+1}{4})$

$r_1=\sqrt{{\left(\frac{-\lambda }{4}\right)}^2+{\left(\frac{{\lambda }^2+1}{4}\right)}^2+10}$

$C_2=(-2,-3),r_2=\sqrt{4+9-3}=\sqrt{10}$

For the circles to cut orthogonally

$C_1{C}_2^2=r_1^2+r_2^2$

$⟹{\left(\frac{8-\lambda }{4}\right)}^2+{\left(\frac{{\lambda }^2+13}{4}\right)}^2=(\frac{{\lambda }^2}{16}+\frac{{\lambda }^2+1}{16}+10)+10$

$⟹{\lambda }^2+64-16\lambda -{\lambda }^2+({\lambda }^2+13{)}^2–(\lambda +1{)}^2=20\times 16$

$⟹24{\lambda }^2-16\lambda -88=0$

$3{\lambda }^2-2\lambda -11=0$

$\lambda =\frac{2\pm \sqrt{4+132}}{6}=\frac{1\pm \sqrt{34}}{3}$

$2$ values of $\lambda$ satisfy two possible circles.