Circles in the coordinate plane | Grade 10

A circle is every point at the same distance from a centre. If you know the distance formula, you already know the equation of a circle: it is the distance formula, written as a rule. This lesson turns the picture "all points $R$ away from $I$ " into algebra.

Equation of a circle

Expanding gives the general form $x^2+y^2-2ax-2by+c=0$ with $c=a^2+b^2-R^2$ . Conversely, an equation of this form is a circle exactly when $R^2=a^2+b^2-c>0$ ; then the centre is $(a, b)$ and the radius is $R=\sqrt{a^2+b^2-c}$ .

Finding a circle

Tangent lines

A line is tangent to a circle when it touches it at exactly one point, and then it is perpendicular to the radius at that point. So the tangent at $M_0$ has normal vector $\overrightarrow{IM_0}$ .

Tangent to a circle at

M

A line is tangent to the circle $(I, R)$ exactly when its distance to the centre equals $R$ : $d(I, \Delta)=R$ . If $d<R$ it cuts the circle in two points; if $d>R$ it misses it.

Key terms

English	Tiếng Việt
circle	đường tròn
centre	tâm
radius	bán kính
diameter	đường kính
tangent line	tiếp tuyến
point of tangency	tiếp điểm
general equation of a circle	phương trình tổng quát của đường tròn
inside / outside	nằm trong / nằm ngoài

Summary

Circle $(I(a, b), R)$ : $(x-a)^2+(y-b)^2=R^2$ .
General form $x^2+y^2-2ax-2by+c=0$ is a circle iff $R^2=a^2+b^2-c>0$ .
Tangent at $M_0$ : normal vector $\overrightarrow{IM_0}$ . A line is tangent iff $d(I, \Delta)=R$ .
A point is inside, on or outside the circle according to $IP^2<R^2$ , $=R^2$ , $>R^2$ .

Practice

Write the circle with centre $(-2, 3)$ and radius $4$ .
Find the centre and radius of $x^2+y^2-4x+6y-3=0$ .
Is $x^2+y^2-2x+4y+6=0$ a circle?
Write the circle with centre $I(2, -1)$ through $A(5, 3)$ .
Write the circle with diameter $AB$ , $A(-1, 3)$ and $B(3, -1)$ .
Find the circle through $O(0, 0)$ , $A(6, 0)$ and $B(0, 8)$ .
Find the tangent to $x^2+y^2=25$ at $M(3, 4)$ .
For which $m$ is $3x+4y+m=0$ tangent to $x^2+y^2=9$ ?
Challenge. Find the common points of $x^2+y^2-4x=0$ and the line $x+y-4=0$ .
Challenge. Find the tangent lines to $x^2+y^2=4$ through $P(4, 0)$ .

Show solutions

$(x+2)^2+(y-3)^2=16$ .
$a=2$ , $b=-3$ , $c=-3$ . $R^2=4+9+3=16$ , so the centre is $(2, -3)$ and $R=4$ .
$a=1$ , $b=-2$ , $c=6$ : $R^2=1+4-6=-1<0$ . Not a circle.
$R^2=IA^2=9+16=25$ . $(x-2)^2+(y+1)^2=25$ .
The centre is $(1, 1)$ and $AB^2=16+16=32$ , so $R^2=\frac{32}{4}=8$ . $(x-1)^2+(y-1)^2=8$ .
$c=0$ ; $A$ : $36-12a=0$ , so $a=3$ ; $B$ : $64-16b=0$ , so $b=4$ . $R^2=9+16=25$ . $(x-3)^2+(y-4)^2=25$ .
$\overrightarrow{OM}=(3, 4)$ : $3(x-3)+4(y-4)=0$ , so $3x+4y-25=0$ .
$\frac{|m|}{5}=3$ , so $m=\pm15$ .
Substitute $y=4-x$ : $x^2+(4-x)^2-4x=0$ , so $2x^2-12x+16=0$ , $x^2-6x+8=0$ , $x=2$ or $x=4$ . Points $(2, 2)$ and $(4, 0)$ .
A line through $P$ : $kx-y-4k=0$ . Tangency: $\frac{|4k|}{\sqrt{k^2+1}}=2$ , so $16k^2=4k^2+4$ , $k^2=\frac13$ , $k=\pm\frac1{\sqrt3}$ . The lines are $x-\sqrt3\,y-4=0$ and $x+\sqrt3\,y-4=0$ .