Question

If $4l^2-5m^2+6l+1=0$ and the line $lx+my+1=0$ touches a fixed circle, then

• A. The center of the circle is at the point $(3,0)$
• B. The radius of the circle is equal to $\sqrt{5}$
• C. The circle passes through origin
• D. None of these

Answer 1

Answer: (A), (B)

The correct options are
A The center of the circle is at the point $(3,0)$
B The radius of the circle is equal to $\sqrt{5}$

Let the circle be ${(x-h)}^2+{(y-k)}^2=r^2$

then $r=\left|\frac{lh+mk+1}{\sqrt{l^2+m^2}}\right|$ $(\because lx+my+1=0$ is tangent $)$

$\Rightarrow r^2(l^2+m^2)={(lh+mk+1)}^2$

$\Rightarrow l^2(h^2-r^2)+m^2(k^2-r^2)+2lmhk+2lh+2mk+1=0$ ...(1)

and $4l^2-5m^2+6l+1=0$ ...(2)

Comparing the coefficients of similar terms

$h^2-r^2=4,k^2-r^2=-5,hk=0.2h=6,2k=0$

$\therefore k=0,h=3$ and $9-r^2=4\Rightarrow r^2=5$

Center $(3,0)$ and radius $=\sqrt{5}$