Exercises — Circle as degenerate conic (e = 0)

(a) Yes. It is $(x-0{)}^2+(y-0{)}^2=7^2$ — the standard form with centre at the origin and radius $7$. The two squared terms have equal coefficient $1$; that equality is the algebra-fingerprint of equal axes, i.e. $e=0$. (b) No — this is an ellipse. Here the number under $x^2$ is $9$ and the number under $y^2$ is $4$; these two numbers are the squared half-widths of the shape in the $x$- and $y$-directions. Because they are different ($9\ne 4$), the shape stretches more one way than the other, so $e\ne 0$. A circle would need these two numbers equal. (c) Yes, provisionally — it matches the general form with $2g=-4,2f=10,c=5$. We only fully confirm a real circle after checking $g^2+f^2-c>0$ (done in L2).

Compare with $(x-h{)}^2+(y-k{)}^2=r^2$. Matching signs carefully: $x-h=x-3\Rightarrow h=3$; and $y-k=y+2=y-(-2)\Rightarrow k=-2$. And $r^2=25\Rightarrow r=5$. Centre $(3,-2)$, radius $5$.

Read coefficients. $2g=-4\Rightarrow g=-2$; $2f=10\Rightarrow f=5$; $c=5$. Why: the general form is $x^2+y^2+2gx+2fy+c=0$, so we just match term by term. Centre $=(-g,-f)=(2,-5)$. Radius $=\sqrt{g^2+f^2-c}=\sqrt{4+25-5}=\sqrt{24}=2\sqrt{6}$. Since $g^2+f^2-c=24>0$, this is a genuine real circle. ✔️

Group $x$-terms and $y$-terms: $(x^2+6x)+(y^2-8y)=0.$ Complete each square. Half of $6$ is $3$, squared is $9$; half of $-8$ is $-4$, squared is $16$. Add both to both sides: $(x^2+6x+9)+(y^2-8y+16)=0+9+16.$ $(x+3{)}^2+(y-4{)}^2=25.$ Centre $(-3,4)$, radius $5$. (See Completing the Square.)

Let $C(-1,5)$ be the centre and $P(x,y)$ be any moving point on the circle. By the definition of a circle, the distance from $C$ to $P$ — written $CP$ — equals the radius $r=3$. Using the Distance Formula and squaring both sides ($CP=r\Rightarrow C{P}^2=r^2$): $(x+1{)}^2+(y-5{)}^2=9.$ Expand: $x^2+2x+1+y^2-10y+25=9$, so $x^2+y^2+2x-10y+17=0.$ Check: $2g=2\Rightarrow g=1$, $2f=-10\Rightarrow f=-5$, centre $(-1,5)$ ✔, radius $\sqrt{1+25-17}=\sqrt{9}=3$ ✔.

The deciding quantity is $g^2+f^2-c$ (the radius squared). (a) $g=-1,f=-2,c=5\Rightarrow g^2+f^2-c=1+4-5=0$. Radius $0$ → a point circle at $(1,2)$. (A degenerate conic — see Degenerate Conics.) (b) $g=1,f=1,c=3\Rightarrow 1+1-3=-1<0$. Radius${}^2$ negative → no real circle (imaginary circle); no real points satisfy it. (c) $g=-3,f=0,c=9\Rightarrow 9+0-9=0$. Another point circle, at $(3,0)$.

The figure below shows all three fates side by side. In the left panel the accent-red curve is a genuine loop (radius${}^2>0$). In the middle panel the loop has shrunk to a single red dot — the point circle, radius${}^2=0$. In the right panel there is no red curve at all: radius${}^2<0$ means not a single real point exists, so nothing can be drawn. Reading left-to-right, you literally watch the radius squeeze from positive, through zero, to impossible.

Use the general form $x^2+y^2+2gx+2fy+c=0$ and substitute each point. $(1,1):1+1+2g+2f+c=0\Rightarrow 2g+2f+c=-2.$ $(2,4):4+16+4g+8f+c=0\Rightarrow 4g+8f+c=-20.$ $(5,3):25+9+10g+6f+c=0\Rightarrow 10g+6f+c=-34.$ Subtract eq1 from eq2: $2g+6f=-18\Rightarrow g+3f=-9.$ Subtract eq1 from eq3: $8g+4f=-32\Rightarrow 2g+f=-8.$ Solve: from the second, $f=-8-2g$. Plug in: $g+3(-8-2g)=-9\Rightarrow g-24-6g=-9\Rightarrow -5g=15\Rightarrow g=-3.$ Then $f=-8-2(-3)=-2.$ And $c=-2-2g-2f=-2+6+4=8.$ $x^2+y^2-6x-4y+8=0.$ Centre $(3,2)$, radius $\sqrt{9+4-8}=\sqrt{5}$. (This uses the $e=0$ fact that the centre is equidistant from all three points.)

First, why is the directrix at $x=a/e$? The focus–directrix definition says every point $P$ on the ellipse obeys $PS=e\cdot PM$, where $S$ is the focus and $PM$ the distance to the directrix line $x=d$. Take the vertex nearest that directrix, at $x=a$: its focus distance is $PS=a-c=a-ae=a(1-e)$, and its directrix distance is $PM=d-a$. So $a(1-e)=e(d-a)$, giving $d-a=\frac{a(1-e)}{e}$, hence $d=a+\frac{a(1-e)}{e}=\frac{a}{e}$. That is where the standard fact $x=a/e$ comes from. (i) $b^2=a^2(1-e^2)=100(1-0.64)=36\Rightarrow b=6.$ Directrix at $x=a/e=10/0.8=\mathrm{12.5.}$ (ii) As $e\to 0$: $b^2=100(1-e^2)\to 100$, so $b\to 10=a$ (axes become equal). Directrix $x=a/e=10/e\to \infty$ (it flees). The two foci $(\pm ae,0)=(\pm 10e,0)\to (0,0)$ merge at the centre. Limiting shape: $\frac{x^2}{100}+\frac{y^2}{100}=1\Rightarrow x^2+y^2=100$ — a circle of radius $10$, $e=0$. (See Ellipse, Eccentricity.)

The figure makes this motion literal. The thin black curves are ellipses at $e=0.8$ and $e=0.5$ (visibly squashed); the bold red curve is the $e=0$ limit — a perfect circle. Watch the two black focus-dots on the $x$-axis slide inward as $e$ shrinks and land together on the red centre-dot at the origin (the red arrow), while the dotted vertical directrix line at $x=12.5$ marches off to the right toward infinity.

A line is tangent when its perpendicular distance from the centre equals the radius. Centre $(0,0)$, radius $\sqrt{8}=2\sqrt{2}$. Rewrite the line as $x-y+k=0$. Distance from $(0,0)$ is $\frac{|k|}{\sqrt{1^2+(-1{)}^2}}=\frac{|k|}{\sqrt{2}}.$ Set equal to radius: $\frac{|k|}{\sqrt{2}}=2\sqrt{2}\Rightarrow |k|=2\sqrt{2}\cdot \sqrt{2}=4.$ So $k=4$ or $k=-4$ (one tangent above, one below — both cases). ✔️

Reflection in $y=x$ swaps coordinates: $(a,b)\mapsto (b,a)$. The radius is unchanged (reflection preserves lengths — this is exactly the total symmetry of $e=0$: no direction is special). Centre of $C_1$ is $(2,3)$, so centre of $C_2$ is $(3,2)$, radius still $4$. $C_2:(x-3{)}^2+(y-2{)}^2=16.$ Distance between centres via Distance Formula: $\sqrt{(3-2{)}^2+(2-3{)}^2}=\sqrt{1+1}=\sqrt{2}.$

The centre is equidistant from $(0,0)$ and $(4,0)$ (the $e=0$ property). Setting the squared distances equal: $h^2+k^2=(h-4{)}^2+k^2\Rightarrow 0=-8h+16\Rightarrow h=2.$ So every such centre lies on the vertical line $x=2$ (the perpendicular bisector of the chord). Centre is $(2,k)$. Now force it through $(2,2)$: distance to $(0,0)$ equals distance to $(2,2)$: $2^2+k^2=(2-2{)}^2+(k-2{)}^2\Rightarrow 4+k^2=k^2-4k+4\Rightarrow 4k=0\Rightarrow k=0.$ Centre $(2,0)$, radius $=\sqrt{2^2+0^2}=2$. Circle: $(x-2{)}^2+y^2=4.$ The figure shows the whole family. The dashed vertical line $x=2$ is the locus of every centre; the thin black circles are two other members (centres at different heights $k$), each still threading both fixed black dots $(0,0)$ and $(4,0)$. The bold red circle is the one member forced to pass through $(2,2)$ — its centre drops exactly onto $(2,0)$ (red arrow), giving radius $2$.

Here $g=-1,f=-1,c=\lambda$, so radius${}^2=g^2+f^2-c=1+1-\lambda =2-\lambda .$ (i) Point circle when radius $=0$: $2-\lambda =0\Rightarrow \lambda =2$ (a single point at $(1,1)$). (ii) Real circle needs radius${}^2>0$: $2-\lambda >0\Rightarrow \lambda <2.$ For $\lambda >2$ the circle is imaginary. (See Degenerate Conics.)