Apply circle formulas for circumference, area, arc length and sector area, and use the standard equation of a circle to find its centre and radius (Geometry).

What this topic is asking

The ACT tests the circle formulas (circumference, area, arc length, sector area) and the standard equation of a circle in the coordinate plane. The formulas are short; the skill is choosing the right one and, for arcs and sectors, scaling by the fraction of the circle the central angle represents.

Circumference and area

The two basic measures of a circle.

Keep them straight by units of thinking: circumference is a length (one power of $r$); area is a region (two powers of $r$).

Arc length and sector area

An arc is part of the circumference; a sector is a "pie slice" of the area. Both are the same fraction of the whole that the central angle is of $360°$.

The same fraction gives arc length: $\frac{72}{360}\cdot 2\pi(10) = \frac{1}{5}\cdot 20\pi = 4\pi$.

The equation of a circle

In the coordinate plane, a circle is the set of points a fixed distance (the radius) from its centre.

So $(x + 1)^{2} + (y - 4)^{2} = 9$ has centre $(-1, 4)$ (because $x + 1 = x - (-1)$) and radius $\sqrt{9} = 3$. Watch the signs: the centre coordinates are the values that make each squared term zero.

Central angles, chords and tangents

A few more circle facts appear on the ACT. A central angle has its vertex at the centre, and it equals the measure of the arc it cuts off, so a $90°$ central angle cuts a quarter-circle arc. A chord is a segment joining two points on the circle; the diameter is the longest chord, passing through the centre. A tangent line touches the circle at exactly one point and is perpendicular to the radius drawn to that point, a fact that turns many tangent problems into right-triangle problems. Recognising that a radius meets a tangent at $90°$ lets you bring in the Pythagorean theorem.

Finding the radius from the equation by completing the square

Sometimes a circle is given in expanded form, such as $x^{2} + y^{2} - 6x + 4y - 12 = 0$, and you must rewrite it in standard form to read the centre and radius. Group and complete the square in $x$ and in $y$: $x^{2} - 6x$ becomes $(x - 3)^{2} - 9$, and $y^{2} + 4y$ becomes $(y + 2)^{2} - 4$. The equation becomes $(x - 3)^{2} + (y + 2)^{2} = 12 + 9 + 4 = 25$, so the centre is $(3, -2)$ and the radius is $5$. The ACT favours equations already in standard form, but knowing how to complete the square handles the harder cases.

Why the fraction method generalises

Treating an arc or sector as a fraction of the whole circle unifies all the partial-circle formulas: whatever fraction $\frac{\theta}{360}$ the angle is, the arc is that fraction of the circumference and the sector is that fraction of the area. You never need separate formulas to memorise; compute the whole, then scale. For the equation of a circle, the reliable habit is to read the centre from the signs inside the squares and to remember the right side is the radius squared.

Try this

Q1. A circle has diameter 14. Find its circumference in terms of $\pi$. [1 point]

• Cue. $C = \pi d = 14\pi$.

Q2. Find the centre and radius of $(x - 2)^{2} + (y - 5)^{2} = 16$. [1 point]

• Cue. Centre $(2, 5)$, radius $\sqrt{16} = 4$.