Question

If $a{l}^2-b{m}^2+2dl+1=0$, where a, b, d are fixed real numbers such that a + b = $d^2$. Then, the line lx + my + 1 = 0 touches a fixed circle

• A. which cuts x-axis orthogonally
• B. with radius equal to b
• C. on which the length of the tangent from origin is $\sqrt{d^2-b}$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A

which cuts x-axis orthogonally

C

on which the length of the tangent from origin is $\sqrt{d^2-b}$

$(d^2-b)l^2+2dl+1=b{m}^2\Rightarrow d^2{l}^2+2dl+1=b(l^2+m^2)\Rightarrow |\frac{dl+1}{\sqrt{l^2+m^2}}|=\left(\sqrt{b}\right)\Rightarrow Center\equiv (d,0)andradius=\sqrt{b}\Rightarrow (x-d{)}^2+(y-0{)}^2=b$