Exercises — General form of circle — converting, finding centre and radius

Match term by term against $x^2+y^2+2gx+2fy+c=0$.

• Coefficient of $x$ is $10$, so $2g=10\Rightarrow g=5$.
• Coefficient of $y$ is $-8$, so $2f=-8\Rightarrow f=-4$.
• Constant is $5$, so $c=5$.

Answer: $g=5,f=-4,c=5$.

• $2g=-4\Rightarrow g=-2$
• $2f=-6\Rightarrow f=-3$
• $c=-3$

Centre $=(-g,-f)=(2,3)$. Radius: first the reality check $g^2+f^2-c=4+9-(-3)=16>0$, so it is real. $r=\sqrt{16}=4.$ Answer: Centre $(2,3)$, radius $4$.

The general form demands the coefficients of $x^2$ and $y^2$ to be exactly $1$. Here they are $2$. So it is not yet in general form.

Divide every term by $2$: $x^2+y^2+2x-4y-3=0.$ Now match:

• $2g=2\Rightarrow g=1$
• $2f=-4\Rightarrow f=-2$
• $c=-3$

Answer: After dividing, $g=1,f=-2,c=-3$.

• $2g=6\Rightarrow g=3$
• $2f=-4\Rightarrow f=-2$
• $c=-12$

Reality check: $g^2+f^2-c=9+4-(-12)=25>0$. Real circle. $\text{Centre}=(-3,2),r=\sqrt{25}=5.$ Answer: Centre $(-3,2)$, radius $5$. (This is exactly the circle drawn in the opening figure.)

Expand each bracket (this reverses Completing the square): $(x-2{)}^2=x^2-4x+4,(y+5{)}^2=y^2+10y+25.$ Add them and set equal to $16$: $x^2+y^2-4x+10y+29=16\Rightarrow x^2+y^2-4x+10y+13=0.$ Confirm: $g=-2,f=5,c=13$. Centre $=(2,-5)$ ✓ (matches $(x-2{)}^2,(y+5{)}^2$). $r=\sqrt{4+25-13}=\sqrt{16}=4$ ✓. Answer: $x^2+y^2-4x+10y+13=0$; centre $(2,-5)$, radius $4$.

The $x^2$ and $y^2$ coefficients are $3$, not $1$ — so divide the whole equation by $3$ first: $x^2+y^2-4x+6y+3=0.$

• $2g=-4\Rightarrow g=-2$
• $2f=6\Rightarrow f=3$
• $c=3$

Reality check: $4+9-3=10>0$. Real. $\text{Centre}=(2,-3),r=\sqrt{10}.$ Answer: Centre $(2,-3)$, radius $\sqrt{10}$.

• $g=1,f=2,c=20$.
• $g^2+f^2-c=1+4-20=-15$.

This is negative, so $r^2<0$ — no real number squares to give it. There are no real points $(x,y)$ satisfying the equation. Answer: Imaginary circle (empty in the real plane).

• $g=-3,f=2,c=13$.
• $g^2+f^2-c=9+4-13=0$.

Since this is exactly $0$, we have $r=0$: a point circle. It has collapsed to its own centre. Centre $=(-g,-f)=(3,-2)$. Answer: Point circle — the single point $(3,-2)$.

Here $g=-2,f=1$, and $c=\lambda$. The reality quantity is $g^2+f^2-c=4+1-\lambda =5-\lambda .$

• Real circle needs $5-\lambda >0\Rightarrow \lambda <5$.
• Point circle needs $5-\lambda =0\Rightarrow \lambda =5$.
• Imaginary when $5-\lambda <0\Rightarrow \lambda >5$.

Answer: (a) $\lambda <5$, (b) $\lambda =5$, (c) $\lambda >5$.

Why the diameter form works (the geometry first). Take any point $P(x,y)$ on the circle other than $A$ or $B$. There is a classic fact — the angle in a semicircle is a right angle (Thales' theorem): if $AB$ is a diameter, then the angle $\angle APB$ is exactly $90^{\circ }$. So $P$ lies on the circle iff the segment $PA$ is perpendicular to the segment $PB$.

Now translate "perpendicular" into algebra. The direction from $P$ to $A$ is the pair $(x_1-x,y_1-y)$, and from $P$ to $B$ is $(x_2-x,y_2-y)$. Two directions are perpendicular exactly when their dot product (the "$\times$ the $x$'s plus $\times$ the $y$'s" test for a right angle) is zero: $(x_1-x)(x_2-x)+(y_1-y)(y_2-y)=0.$ Flipping the sign of both factors in each product changes nothing, so this is the same as $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0.$ That is the diameter form (see Equation of circle from endpoints of diameter) — not a formula to memorise, but the direct statement "the angle at $P$ is a right angle."

Substitute $A(1,2)$ and $B(5,8)$: $(x-1)(x-5)+(y-2)(y-8)=0.$ Expand: $x^2-6x+5+y^2-10y+16=0\Rightarrow x^2+y^2-6x-10y+21=0.$ Answer: $x^2+y^2-6x-10y+21=0$. (Check: centre $=(3,5)$, the midpoint of $A$ and $B$ ✓; $r=\sqrt{9+25-21}=\sqrt{13}$, which equals half of $|AB|=\sqrt{16+36}=\sqrt{52}=2\sqrt{13}$ ✓.)

The radius is the distance from centre to the given point on the circle: $r=\sqrt{(0-(-3){)}^2+(0-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5.$ Start in Standard form of circle: $(x+3{)}^2+(y-4{)}^2=25.$ Expand to general form: $x^2+6x+9+y^2-8y+16=25\Rightarrow x^2+y^2+6x-8y=0.$ Note the constant $c=0$; this is the fingerprint of "passes through the origin" (put $x=y=0$ and the equation holds only if $c=0$). Answer: $x^2+y^2+6x-8y=0$.

Let the circle be $x^2+y^2+2gx+2fy+c=0$. Plug in each point.

Point $(0,0)$: $0+0+0+0+c=0\Rightarrow c=0$.

Point $(4,0)$: $16+0+8g+0+c=0$. With $c=0$: $16+8g=0\Rightarrow g=-2$.

Point $(0,6)$: $0+36+0+12f+c=0$. With $c=0$: $36+12f=0\Rightarrow f=-3$.

So the equation is $x^2+y^2-4x-6y=0.$ Centre $=(-g,-f)=(2,3)$. Radius $=\sqrt{4+9-0}=\sqrt{13}$. Answer: $x^2+y^2-4x-6y=0$; centre $(2,3)$, radius $\sqrt{13}$.

Get the geometry first. Here $g=-2,f=-3,c=-3$. Centre $=(-g,-f)=(2,3)$, radius $=\sqrt{g^2+f^2-c}=\sqrt{4+9-(-3)}=\sqrt{16}=4$.

The tangency test. A circle is tangent to a straight line when the perpendicular distance from its centre to the line equals its radius (see Tangent to a circle). The $x$-axis is the line $y=0$, and the perpendicular distance from the centre $(2,3)$ straight down to it is simply the centre's height $|k|=3$.

Here distance $=3$ but radius $=4$, so distance $<$ radius: the circle crosses the $x$-axis in two points rather than touching it. So this equation is not tangent to the $x$-axis — it is a secant.

Finding the tangent version. Tangency to $y=0$ requires distance $=$ radius, i.e. height $|k|=$ radius. With centre $(2,3)$ the radius must be exactly $3$, so we need $g^2+f^2-c=9$, i.e. $4+9-c=9\Rightarrow c=4$. The tangent circle is therefore $x^2+y^2-4x-6y+4=0.$ Point of contact. For a tangent circle sitting above the $x$-axis, the touch point lies directly below the centre, at the same $x$-value with $y=0$: the point is $(2,0)$.

Answer: The given circle ($c=-3$) is not tangent — it is a secant (distance $3<$ radius $4$). Adjusting the constant to $c=4$ gives the tangent circle $x^2+y^2-4x-6y+4=0$, which touches the $x$-axis at $(2,0)$.

Centre on $y=x$: the centre is $(-g,-f)$, so it satisfies $y=x$ when $-f=-g\Rightarrow f=g$. …(I)

Through $(1,0)$: substitute $x=1,y=0$: $1+0+2g+0+c=0\Rightarrow 1+2g+c=0$. …(II)

Through $(0,1)$: substitute $x=0,y=1$: $0+1+0+2f+c=0\Rightarrow 1+2f+c=0$. …(III)

Subtracting (III) from (II): $2g-2f=0$, i.e. $f=g$ — the same condition as (I), so the two points are automatically symmetric about $y=x$ and give no new information. We are left with two independent facts, $f=g$ and $c=-1-2g$, but three unknowns — so the answer is a one-parameter family of circles (this is exactly a Family of circles threaded through the two fixed points).

Parametrisation. Choosing $g$ as the free parameter: $f=g,c=-1-2g.$

Show every such circle is real. We need the reality quantity positive: $g^2+f^2-c=g^2+g^2-(-1-2g)=2g^2+2g+1.$ Treat this as a quadratic in $g$. Its discriminant is $2^2-4\cdot 2\cdot 1=4-8=-4<0$, and its leading coefficient $2>0$, so the expression is strictly positive for every real $g$. Hence $g^2+f^2-c>0$ always — every circle in the family is a genuine real circle.

Answer: $f=g$ and $c=-1-2g$ for any real $g$; since $2g^2+2g+1>0$ for all $g$, every such circle is real (there is no imaginary or point-circle case).

Matching (see Conic sections general equation): $A=1,B=0,C=1,D=-8,E=6,F=0.$ A general second-degree conic is a circle exactly when $A=C(\text{equal squared coefficients})\text{and}B=0(\text{no }xy\text{ cross term}).$ Both hold: $A=C=1$ and $B=0$. ✓

Now read the circle data. Since $A=C=1$, we have $2g=D=-8\Rightarrow g=-4$ and $2f=E=6\Rightarrow f=3$, with $c=F=0$. Centre $=(4,-3)$; radius $=\sqrt{16+9-0}=\sqrt{25}=5$. Answer: $A=C=1$, $B=0$ (the circle conditions); centre $(4,-3)$, radius $5$.