Find circumference and area of circles, compute arc length and sector area as fractions of the whole, and apply the central-angle and inscribed-angle relationships.

What this topic is asking

The Geometry category covers circles (the G-C and G-GMD standards). On the Grade 10 MCAS you find circumference and area, compute arc length and sector area as fractions of the whole circle, and apply the central-angle and inscribed-angle relationships. The circle area and circumference formulas are on the reference sheet, so the skill is in using the fraction of the circle and the angle relationships correctly.

Circumference and area

The two whole-circle measures, both provided on the reference sheet, are:

where $r$ is the radius and $d = 2r$ the diameter. The most common slip is mixing them up: circumference is linear in $r$ (a perimeter), while area is quadratic in $r$ (it has $r^2$). For $r = 6$, the circumference is $12\pi$ but the area is $36\pi$.

Answers are often requested in terms of $\pi$ (exact) rather than as a decimal, so leave the $\pi$ in place unless a decimal is asked for.

Arc length and sector area

An arc is part of the circle's edge, and a sector is a pie-slice region. Both are the same fraction of the whole circle as the central angle is of $360^\circ$:

The single idea to hold onto: find what fraction of the full circle the angle represents, then take that fraction of the circumference (for an arc) or the area (for a sector). A semicircle is $\frac{180}{360} = \frac{1}{2}$; a quarter circle is $\frac{90}{360} = \frac{1}{4}$.

Central and inscribed angles

Two angle relationships appear on the MCAS:

• A central angle (vertex at the center) has the same measure as the arc it intercepts. A $70^\circ$ central angle cuts a $70^\circ$ arc.
• An inscribed angle (vertex on the circle) is half the central angle that intercepts the same arc, and therefore half the intercepted arc. So an inscribed angle intercepting a $70^\circ$ arc measures $35^\circ$.

A special case worth knowing: an inscribed angle that intercepts a semicircle (a diameter) is a right angle ($90^\circ$), because the arc is $180^\circ$ and half of that is $90^\circ$.

Composite and partial regions

The MCAS sometimes asks for the area of a region built from circle parts, such as a semicircle on top of a rectangle, or the area left when a circle is cut from a square. The method is to find each piece and add or subtract. For a running track with two semicircular ends, the two semicircles together make one full circle, so their combined area is $\pi r^2$. For a circular hole of radius 2 cut from a square of side 6, the remaining area is $36 - \pi(2)^2 = 36 - 4\pi$ square units. Keeping the $\pi$ exact and combining the pieces carefully is the key.

Equations of circles on the coordinate plane

A circle centered at $(h, k)$ with radius $r$ has the equation $(x - h)^2 + (y - k)^2 = r^2$. Reading it backwards, the equation $(x - 2)^2 + (y + 3)^2 = 25$ describes a circle with center $(2, -3)$ and radius $\sqrt{25} = 5$. Watch the sign flip on the center, just as with vertex form for parabolas: $(y + 3)$ gives a $k$ of $-3$. This connects the circle to coordinate geometry and the distance formula, since every point on the circle is exactly $r$ from the center.

Try this

Q1. A circle has diameter 14. What is its circumference, in terms of $\pi$?

• Cue. $C = \pi d = 14\pi$.

Q2. An inscribed angle intercepts an arc of $100^\circ$. What is the inscribed angle?

• Cue. Half the arc: $50^\circ$.