Coordinate geometry of the circle and radians

The equation of a circle

A point $(x, y)$ is on the circle with centre $(h, k)$ and radius $r$ when its distance from the centre is $r$. The distance formula gives $\sqrt{(x-h)^2 + (y-k)^2} = r$. Squaring both sides gives the equation.

Expanding the brackets gives the general form (this is the form in Formulae and Tables):

Comparing the two forms: the centre is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$. The signs of $g$ and $f$ are the opposite of the signs in the brackets, which is why the centre has the minus signs.

Finding the equation of a circle

To write the equation you need the centre and the radius. Most problems give them to you in disguise.

Diameter given. The centre is the midpoint of the diameter, and the radius is half its length (or the distance from the centre to either endpoint).

Three points given. Substitute each point into $x^2 + y^2 + 2gx + 2fy + c = 0$. This gives three linear equations in $g$, $f$ and $c$.

Lines, chords and tangents

To find where a line meets a circle, substitute the line into the circle's equation (write $y$ in terms of $x$, or the reverse). You get a quadratic:

• two solutions: the line cuts the circle at two points (a chord lies between them);

• one repeated solution (discriminant $= 0$): the line is a tangent;

• no real solutions: the line misses the circle.

So the path does get wet, over about $8.49$ m.

Tangents

A tangent touches the circle at exactly one point, and it is perpendicular to the radius at that point.

To find the equation of the tangent at a point $P$ on the circle:

1. Find the slope of the radius from the centre to $P$.

2. The tangent has the negative reciprocal slope.

3. Use $y - y_1 = m(x - x_1)$ with the point $P$.

Tangents from an external point. If $Q$ is outside a circle with centre $C$ and radius $r$, the radius at the point of contact is perpendicular to the tangent, so Pythagoras gives

For the circle above and $Q(9, 2)$: $|QC| = 8$, so the tangent length is $\sqrt{64 - 25} = \sqrt{39}$.

Parallel tangents. A line with a given slope $m$ is a tangent when its distance from the centre equals $r$. Write the line as $ax + by + c = 0$ with $c$ unknown, set the distance formula equal to $r$, and solve for $c$. This gives two answers, one on each side of the circle.

Radians, arcs and sectors

Degrees split a full turn into $360$ equal parts, which is an arbitrary choice. Radians use the circle itself: one radian is the angle at the centre when the arc has the same length as the radius.

Since the circumference is $2\pi r$, a full turn is $2\pi$ radians.

When $\theta$ is in radians, the arc is the fraction $\frac{\theta}{2\pi}$ of the circumference, and the sector is the same fraction of the area:

Segment. A segment is the region between a chord and its arc. Take the sector and subtract the triangle formed by the two radii and the chord. The triangle has area $\frac{1}{2}r^2\sin\theta$, so

Summary

• A circle with centre $(h, k)$ and radius $r$ has equation $(x-h)^2 + (y-k)^2 = r^2$. In general form $x^2 + y^2 + 2gx + 2fy + c = 0$ the centre is $(-g, -f)$ and $r = \sqrt{g^2 + f^2 - c}$.

• For a diameter, the centre is the midpoint. For three points, substitute each into the general form and solve for $g$, $f$, $c$. A point is inside, on or outside depending on whether its squared distance from the centre is less than, equal to or greater than $r^2$.

• Substitute a line into the circle to find intersections. Two roots give a chord, a repeated root gives a tangent.

• A tangent is perpendicular to the radius at the point of contact. Tangent length from an external point $= \sqrt{|QC|^2 - r^2}$.

• $\pi$ rad $= 180^\circ$. With $\theta$ in radians, $l = r\theta$, $A = \frac{1}{2}r^2\theta$, and segment area $= \frac{1}{2}r^2(\theta - \sin\theta)$.