The Circle — Centre (0, 0)

Section 1 of 3

Equation & Radius

A circle centred at the origin has one simple equation. Everything in Part 1 starts here.

Centre (0, 0)

$x^2 + y^2 = r^2$. Compare to $x^2 + y^2 = $ a number — that number is $r^2$, not $r$. So take the square root to get the radius.

Worked example — find the radius

Find the radius of each: (i) $x^2+y^2=25$; (ii) $x^2+y^2=49$; (iii) $x^2+y^2=12$.

(i) $r^2 = 25 \Rightarrow r = 5$

(ii) $r^2 = 49 \Rightarrow r = 7$

(iii) $r^2 = 12 \Rightarrow r = \sqrt{12} = 2\sqrt3$

$5,\ 7,\ 2\sqrt3$

Worked example — given the radius

Find the equation, centre $(0,0)$, with radius $8$, then radius $10$.

$x^2 + y^2 = 8^2 = 64$

$x^2 + y^2 = 10^2 = 100$

$x^2+y^2=64$ and $x^2+y^2=100$

You try

Find the equation, centre $(0,0)$, with radius $\sqrt5$, then radius $\sqrt{11}$.

Squaring a surd kills the root: $(\sqrt{n})^2 = n$.

$x^2 + y^2 = (\sqrt5)^2 = 5$

$x^2 + y^2 = (\sqrt{11})^2 = 11$

$x^2+y^2=5$ and $x^2+y^2=11$

Section 2 of 3

Given a Point on the Circle

Radius = distance

The radius is the distance from the centre to the point. You only need $r^2$, so skip the square root: $r^2 = (x_2-x_1)^2 + (y_2-y_1)^2$.

Worked example — a point

Find the circle, centre $(0,0)$, through $(1, 3)$.

$r^2 = (1-0)^2 + (3-0)^2 = 1 + 9 = 10$

$x^2 + y^2 = 10$

Worked example — another

Find the circle, centre $(0,0)$, through $(-1, 9)$.

$r^2 = (-1)^2 + 9^2 = 1 + 81 = 82$

$x^2 + y^2 = 82$

You try

Find the circle, centre $(0,0)$, through $(3, 7)$.

$r^2 = 3^2 + 7^2$. No need for the square root.

$r^2 = 9 + 49 = 58$

$x^2 + y^2 = 58$

Section 3 of 3

Intersection & the Axes

Line meets circle

Make $y$ the subject of the line, substitute into the circle, tidy to a quadratic in $x$, solve. Two solutions = a chord; one = a tangent; none = a miss.

Worked example — a chord

Find where $x + y = 3$ meets $x^2 + y^2 = 5$.

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

$2x^2 - 6x + 4 = 0 \Rightarrow x^2 - 3x + 2 = 0 \Rightarrow (x-2)(x-1)=0$

$x=2 \Rightarrow y=1$; $x=1 \Rightarrow y=2$

$(2, 1)$ and $(1, 2)$

Worked example — a tangent

Find where $x + y = 4$ meets $x^2 + y^2 = 8$.

$y = 4 - x$, so $2x^2 - 8x + 8 = 0 \Rightarrow (x-2)^2 = 0$

$x = 2$ (twice), $y = 2$ — one point, so the line is a tangent

$(2, 2)$

Cutting the axes

The $x$-axis means $y = 0$; the $y$-axis means $x = 0$. Substitute and solve.

Worked example — cuts the x-axis

Find where $x^2 + y^2 = 9$ cuts the $x$-axis.

$y = 0$: $x^2 = 9 \Rightarrow x = \pm 3$

$(-3, 0)$ and $(3, 0)$

You try

Find where $x + y = 7$ meets $x^2 + y^2 = 25$.

$y = 7 - x$; substitute and solve the quadratic in $x$.

$2x^2 - 14x + 24 = 0 \Rightarrow x^2 - 7x + 12 = 0 \Rightarrow (x-3)(x-4)=0$

$x=3 \Rightarrow y=4$; $x=4 \Rightarrow y=3$

$(3, 4)$ and $(4, 3)$

That’s Part 1.

The circle centred at the origin: its equation, finding the radius, using a point, line intersections (chord/tangent/miss), and cutting the axes. Part 2: tangents and the position of a point.