The Circle — Centre (h, k)

Section 1 of 3

Centre (h, k) — Equation & Reading Off

Same idea as Part 1, but the centre can be anywhere, not just the origin.

Centre (h, k)

$(x - h)^2 + (y - k)^2 = r^2$. Subtract the centre coordinates inside the brackets. Watch the signs: centre $(3, -2)$ gives $(x-3)^2 + (y+2)^2$; if you see $(x+5)^2$, the $x$-coordinate of the centre is $-5$.

Worked example — write the equation

Circle with centre $(3, 2)$, radius $5$; and centre $(3, -2)$, radius $8$.

$(x-3)^2 + (y-2)^2 = 5^2 = 25$

$(x-3)^2 + (y+2)^2 = 8^2 = 64$

$(x-3)^2+(y-2)^2=25$ and $(x-3)^2+(y+2)^2=64$

Worked example — read off centre & radius

Find the centre and radius of $(x-1)^2 + (y-3)^2 = 36$ and $(x-3)^2 + (y+5)^2 = 49$.

$(x-1)^2+(y-3)^2=36$: centre $(1, 3)$, $r = \sqrt{36} = 6$

$(y+5)$ is $(y-(-5))$, so $k = -5$: centre $(3, -5)$, $r = 7$

$(1,3),\ r=6$ and $(3,-5),\ r=7$

You try

Write the equation of the circle with centre $(-1, -6)$, radius $\sqrt3$.

Negative centre coordinates flip to plus signs inside the brackets.

$(x-(-1))^2 + (y-(-6))^2 = (\sqrt3)^2$

$(x+1)^2 + (y+6)^2 = 3$

Section 2 of 3

Given a Point & Diameters

Radius = distance

The radius is the distance from the centre to a point on the circle: $r^2 = (x_2-x_1)^2 + (y_2-y_1)^2$. Then write $(x-h)^2 + (y-k)^2 = r^2$.

Worked example — centre and a point

Circle with centre $(3, -1)$ through $(1, 4)$.

$r^2 = (1-3)^2 + (4-(-1))^2 = (-2)^2 + 5^2 = 4 + 25 = 29$

$(x-3)^2 + (y+1)^2 = 29$

Worked example — endpoints of a diameter

$(2, -1)$ and $(6, 7)$ are the endpoints of a diameter. Find the circle.

Centre = midpoint $= \left(\tfrac{2+6}{2}, \tfrac{-1+7}{2}\right) = (4, 3)$

$r^2 = (2-4)^2 + (-1-3)^2 = 4 + 16 = 20$

$(x-4)^2 + (y-3)^2 = 20$

Worked example — the other endpoint

$(2, 1)$ is one endpoint of a diameter of $(x-3)^2 + (y+2)^2 = 25$. Find the other endpoint.

Centre $(3, -2)$ is the midpoint. Let the other be $(a, b)$

$\tfrac{2+a}{2} = 3 \Rightarrow a = 4$; $\tfrac{1+b}{2} = -2 \Rightarrow b = -5$

$(4, -5)$

You try

$(4, -1)$ and $(-6, 3)$ are the endpoints of a diameter. Find the circle.

Centre = midpoint of the two endpoints; $r^2$ = distance$^2$ from the centre to one endpoint.

Centre $= \left(\tfrac{4+(-6)}{2}, \tfrac{-1+3}{2}\right) = (-1, 1)$

$r^2 = (4-(-1))^2 + (-1-1)^2 = 25 + 4 = 29$

$(x+1)^2 + (y-1)^2 = 29$

Section 3 of 3

Symmetry & Translation

Under both central symmetry and a translation, the radius never changes — only the centre moves. So $r^2$ on the right stays the same.

Central symmetry in a point

The centre of symmetry $(p, q)$ is the midpoint of the old centre and the new one. Solve $\tfrac{h+h'}{2} = p$ and $\tfrac{k+k'}{2} = q$ for the new centre.

Worked example — central symmetry

Find the image of $(x-1)^2 + (y+2)^2 = 81$ under central symmetry in $(3, -5)$.

Old centre $(1, -2)$. $\tfrac{1+h'}{2} = 3 \Rightarrow h' = 5$; $\tfrac{-2+k'}{2} = -5 \Rightarrow k' = -8$

New centre $(5, -8)$, radius unchanged

$(x-5)^2 + (y+8)^2 = 81$

Translation

A translation slides everything by a fixed shift. Read the shift from where one point goes (e.g. $(2,5) \to (1,7)$ is $x: -1,\ y: +2$), then apply the same shift to the centre.

Worked example — translation

Find the image of $(x-3)^2 + (y+1)^2 = 25$ under the translation $(2, 5) \to (1, 7)$.

Shift: $x: -1,\ y: +2$

Centre $(3, -1) \to (3-1,\ -1+2) = (2, 1)$

$(x-2)^2 + (y-1)^2 = 25$

You try

Find the image of $(x+5)^2 + (y+3)^2 = 36$ under the translation $(2, -1) \to (4, -6)$.

Shift is $x: +2,\ y: -5$. Apply it to the centre $(-5, -3)$; radius unchanged.

Shift: $x: +2,\ y: -5$

Centre $(-5, -3) \to (-5+2,\ -3-5) = (-3, -8)$

$(x+3)^2 + (y+8)^2 = 36$

That’s The Circle — the whole topic.

Circles at the origin and at any $(h, k)$: equations, radius, points, tangents, position, diameters, symmetry and translation. That’s the full Ordinary Level Circle course.