If $4\alpha^2–5\beta^2+6\alpha+1=0$.Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. If $4\alpha^2–5\beta^2+6\alpha+1=0$. Prove that $x\alpha+y\beta+1=0$touches a Definite circle. Find the centre and radius of the circle. I tried to solve this question by taking a General equation of circle and then substituting the values but could not proceed further I took the line as a tangent and try to prove it by equating the radius with perpendicular distance of the line from the assumed centre.

We have $\beta(\alpha)=\pm\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}$ so you want to find the envelope of the family $f_\alpha(x,y)=x\alpha+y\beta(\alpha)+1=0$.
In other words, $F(x,y,\alpha):=x\alpha\pm y\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}+1=0$, $\dfrac{\partial}{\partial\alpha}F(x,y,\alpha)=0$,
i.e.,
$$
\left\{
\begin{aligned}
x\alpha\pm y\sqrt{\dfrac{4\alpha^2+6\alpha+1}{5}}+1&=0\\
x\pm y\dfrac{4\alpha+3}{\sqrt{4\alpha^2+6\alpha+1}\sqrt{5}} &=0
\end{aligned}
\right.
$$
which gives
$$
x=\frac{4\alpha+3}{3\alpha+1}, y=\mp\frac{\sqrt{5}\sqrt{4\alpha^2+6\alpha+1}}{3\alpha+1}
$$
and eliminating $\alpha$ gives $x^2+y^2-6x+4=0$, from which you can read off the centre and radius of the circle.
