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 $$m=0\text{ and }s^2-4p=t^2-4q>0\,.$$

  • $\begingroup$ 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. $\endgroup$ – amd Apr 21 at 6:41

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