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.
 A: 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.
A: Suppose that the set $S_\Gamma\subseteq\mathbb{R}^2$ of straight lines of the form $\big\{(x,y)\in\mathbb{R}^2\,\big|\,ax+by+1=0\big\}$, where $a$ and $b$ are real parameters, that are tangent to a single circle $\Gamma$.  Let $(h,k)$ be the center of $\Gamma$, and $r$ its radius.  Then, $b(x-h)=a(y-k)$ is the line connecting $(h,k)$ to the point of tangency with the line $a x+b y+1=0$ in $S_\Gamma$.  The tangent point is then
$$(x,y)=\left(h-\frac{a\big(ah+bk+1\big)}{a^2+b^2},k-\frac{b\big(ah+bk+1\big)}{a^2+b^2}\right)\,.$$
That is,
$$\begin{align}r^2&=\left(\frac{a\big(ah+bk+1\big)}{a^2+b^2}\right)^2+\left(\frac{b\big(ah+bk+1\big)}{a^2+b^2}\right)^2\\&=\frac{(ah+bk+1)^2}{a^2+b^2}\,.\end{align}$$
That is,
$$(h^2-r^2)a^2+2hkab+(k^2-r^2)b^2+2ha+2kb+1=0\,.\tag{*}$$
This equation parametrizes all straight lines in $S_\Gamma$.
In this problem, $a=\alpha$ and $b=\beta$.  Since the equation relating $\alpha$ and $\beta$ is $$4\alpha^2-5\beta^2+6\alpha+1=0\,,$$ we conclude by comparing the above equation with (*) that
$$h^2-r^2=4\,,\,\,k^2-r^2=-5\,,\,\,2h=6\,,\text{ and }2k=0\,.$$
This gives $h=3$, $k=0$, and $r=\sqrt{5}$, in agreement with user10354138's answer.
Conversely, if,  for some circle $\Gamma$, the equation
$$pa^2+mab+qb^2+sa+tb+1=0$$
parametrizes straight lines in $S_\Gamma$ with the equation $ax+by+1=0$, then it must hold that
$$s^2-4p=t^2-4q>0\,.$$
When this happens, $h=\dfrac{s}{2}$, $k=\dfrac{t}{2}$, and $r=\dfrac{\sqrt{s^2-4p}}{2}=\dfrac{\sqrt{t^2-4q}}{2}$.  Hence, we have the following result.

Proposition.  Let $m,p,q,s,t\in\mathbb{R}$ be such that $$\left\{(x,y)\in\mathbb{R}^2\,\big|\,px^2+mxy+qy^2+sx+ty+1=0\right\}$$ is a conic section with infinitely many points (i.e., it is nonempty and contains more than one point).  There exists a circle $\Gamma$ tangent to all the straight lines of the form $$\big\{(x,y)\in\mathbb{R}^2\,\big|\,ax+by+1=0\big\}$$ whose parameters $a\in\mathbb{R}$ and $b\in\mathbb{R}$ satisfy $$pa^2+mab+qb^2+sa+tb+1=0$$ if and only if
  $$s^2-4p=t^2-4q>0\,.$$

A: Setting $\alpha=0$ gives $\beta=\pm\frac1{\sqrt5}$, so the horizontal lines $y=\pm\sqrt5$ are tangents. Similarly, setting $\beta=0$ yields $\alpha=\frac14(-3\pm\sqrt5)$, which means that the vertical lines $x=3\pm\sqrt 5$ are also tangents. These four lines define a square, so they are all tangent to a circle centered at $(3,0)$ with radius equal to $\sqrt5$.  
This is the only possible circle to which all the lines are tangent, but we still need to check that they are actually tangent to it. To do so, we examine the dual conic equation to that of the circle:$$\begin{bmatrix}\alpha&\beta&1\end{bmatrix} \begin{bmatrix}1&0&-3\\0&1&0\\-3&0&4\end{bmatrix}^{-1} \begin{bmatrix}\alpha\\\beta\\1\end{bmatrix} = -\frac15(4\alpha^2-5\beta^2+6\alpha+1)=0,$$ which agrees with the original constraint on $\alpha$ and $\beta$.  
In fact, using the idea of dual conics we could have found the equation of the circle directly: $$\begin{bmatrix}4&0&3\\0&-5&0\\3&0&1\end{bmatrix}^{-1} = -\frac15\begin{bmatrix}1&0&-3\\0&1&0\\-3&0&4\end{bmatrix},$$ which corresponds to the circle $x^2+y^2-6x+4=0$.
