Question:

Step 1: The line $ax + by + 1 = 0$ touches a fixed circle for all values of $a$ and $b$ satisfying $a^2 - 3b^2 + 4a + 1 = 0$. The distance from the centre $(h, k)$ of the circle to the line equals the radius $r$: $$\frac{|ah + bk + 1|}{\sqrt{a^2 + b^2}} = r$$ $$\Rightarrow (ah + bk + 1)^2 = r^2(a^2 + b^2)$$ Step 2: Expand: $$a^2h^2 + b^2k^2 + 1 + 2abhk + 2ah + 2bk = r^2 a^2 + r^2 b^2$$ $$a^2(h^2 - r^2) + b^2(k^2 - r^2) + 2abhk + 2ah + 2bk + 1 = 0$$ Step 3: This must be consistent with $a^2 - 3b^2 + 4a + 1 = 0$, i.e., $a^2 + 4a + 1 = 3b^2$. Compare coefficients by matching the constraint. From the constraint: coefficient of $a^2$ is 1, coefficient of $b^2$ is $-3$, coefficient of $a$ is 4, constant is 1, and $b^2$ coefficient relative to $a^2$ is $-3$. Step 4: Match ratios: $$\frac{h^2 - r^2}{1} = \frac{k^2 - r^2}{-3} = \frac{2h}{4} = \frac{2k}{0} = \frac{1}{1}$$ From $\frac{2k}{0}$: $k = 0$. From $\frac{2h}{4} = 1$: $h = 2$. From $h^2 - r^2 = 1$: $4 - r^2 = 1 \Rightarrow r^2 = 3 \Rightarrow r = \sqrt{3}$. Check: $\frac{k^2 - r^2}{-3} = \frac{0 - 3}{-3} = 1$ ✓. Step 5: Centre is $(2, 0)$ and radius is $\sqrt{3}$. Options (a) and (b) are correct. Step 6: Check option (c): Distance from $(2,0)$ to $(2,3)$ is $3 \neq \sqrt{3}$, so $(2,3)$ is not on the circle. Option (c) is incorrect.