Question

[Key Concept: The constraint $4a^{2} - 5b^{2} + 6a + 1 = 0$ with $ax + by + 1 = 0$ tangent to a fixed circle: matching the tangency condition $\frac{|a\ x_{0} + b\ y_{0} + 1|}{\sqrt{a^{2} + b^{2}}} = r$ fixes the centre and radius.]

$$9a^{2} + 6a + 1 = 5\left( a^{2} + b^{2} \right)$$

$$\Rightarrow \frac{3a + 1}{\sqrt{a^{2} + b^{2}}} = \sqrt{5} $$$\Rightarrow \bot$ ar distance of $(3,0)$

from $ax + by + 1 = 0$ is $\sqrt{5}$

⇒ Centre of circle $= (3,0)$

$$\text{~}\text{radius}\text{~} = \sqrt{5}$$