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.

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, $$(h^2-r^2)a^2+(k^2-r^2)b^2+2ha+2kb+1=0\,.\tag{*}$$ 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 $$pa^2+qb^2+sa+tb+1=0$$ 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 $$m=0\text{ and }s^2-4p=t^2-4q>0\,.$$

• Your equation (*) is missing the term $2hkab$. Fortunately, $2hk=0$ is consistent with the other equations you derive by comparing terms. I’ll also note that the expression you derive for $r^2$ can be obtained directly from the point-line distance formula. – amd Apr 21 at 6:41