Circles

GE.G.1Central angles & arcs01/04

Central angles and arcs.

Step01of 04

A central angle has its vertex AT the center of the circle. Its two sides are radii. The arc between those radii has a measure EQUAL to the central angle.

Central angle = arc measure (in degrees).

GE.G.1 — Central angles and arcs.

Step 01 of 04

A central angle has its vertex AT the center of the circle. Its two sides are radii. The arc between those radii has a measure EQUAL to the central angle.

Central angle = arc measure (in degrees).

Step 02 of 04

Major vs. minor arc. Two points on a circle divide it into two arcs.

Minor arc	→less than 180°, named with the two endpoints (arc AB)
Major arc	→more than 180°, named with three points (arc ACB) to disambiguate
Semicircle	→exactly 180°, formed by a diameter

The two arcs together always sum to $360°$ — the full circle.

Step 03 of 04

Congruent chords ↔ congruent arcs. In the same circle (or in congruent circles):

Congruent chords	→cut off congruent arcs
Congruent arcs	→are subtended by congruent chords
Equal central angles	→cut off equal arcs (and equal chords)

All three conditions are equivalent — equal in any one of these measurements means equal in all of them.

Step 04 of 04

Worked example. A circle has central angle $∠ A O B = 75°$ . The chord $A B$ cuts the circle into two arcs.

minor arc AB = 75°, major arc = 360° - 75° = 285°

Key insight

Central angle = arc measure (in degrees). Minor arc < 180°, major arc > 180°, full circle = 360°. Equal central angles ↔ equal arcs ↔ equal chords — all three travel together.

GE.G.2 — Inscribed angles.

Step 01 of 04

An inscribed angle has its vertex ON the circle (not at the center). Its two sides are chords. The arc it intercepts has a measure twice the inscribed angle.

inscribed angle = \frac{1}{2} \cdot intercepted arc

Inscribed angle = ½ × intercepted arc.

Step 02 of 04

Worked example. Inscribed angle $∠ P = 35°$ . Find the intercepted arc $A B$ .

arc AB = 2 \cdot 35° = 70°

And going the other direction: if intercepted arc is $110°$ , the inscribed angle is $55°$ .

Step 03 of 04

Two consequences worth memorizing.

Inscribed angles intercepting the SAME arc	→are CONGRUENT (all equal half the same arc)
Inscribed angle in a SEMICIRCLE	→is a RIGHT angle (intercepts a 180° arc, half = 90°)
Opposite angles of a quadrilateral inscribed in a circle	→are SUPPLEMENTARY (sum to 180°)

The "right angle in a semicircle" is one of the most-used facts in geometry — it lets you spot right triangles inside circles instantly.

Step 04 of 04

Quick comparison: central vs. inscribed.

Central angle	→vertex at CENTER, equals the arc
Inscribed angle	→vertex ON the circle, equals HALF the arc

If both angles intercept the same arc, the central is exactly twice the inscribed.

Key insight

Inscribed angle = half the intercepted arc. The triangle inscribed in a semicircle is automatically right (angle opposite diameter = 90°). Two inscribed angles on the same arc are equal — same arc, same half.

GE.G.3 — Tangent, secant, and chord relationships.

Step 01 of 04

Three lines that interact with a circle.

Chord	→segment with both endpoints on the circle
Secant	→line that crosses the circle at TWO points
Tangent	→line that touches the circle at EXACTLY ONE point

The diameter is the longest chord, and a chord extended forms a secant.

Step 02 of 04

Tangent ⟂ radius. A radius drawn to the point of tangency is ALWAYS perpendicular to the tangent line. This is the most-used circle fact in proofs.

Two tangent segments from the same external point are CONGRUENT. Useful for finding distances from external points.

Step 03 of 04

Angle formed by two chords intersecting INSIDE the circle.

angle = \frac{1}{2} (arc 1 + arc 2)

The intercepted arcs are the two opposite the angle.

Angle formed by two secants/tangents from a point OUTSIDE.

angle = \frac{1}{2} (far arc - near arc)

Inside → average the arcs (sum / 2). Outside → difference of arcs / 2.

Step 04 of 04

Segment-product theorems for SEGMENT lengths.

Two chords intersect inside	→ $a \cdot b = c \cdot d$ (each chord is split into two pieces; the products are equal)
Two secants from external point	→ $a (a + b) = c (c + d)$ (whole-times-outside, both equal)
Tangent and secant from external point	→ $t^{2} = a (a + b)$ (tangent squared = whole-times-outside of secant)

Key insight

Tangent ⟂ radius is the #1 fact. For ANGLES: inside chords average the arcs, outside secants/tangents take the half-difference. For LENGTHS: the product rules — chord-chord, secant-secant, tangent-secant — all follow the "products of pieces" pattern.

GE.G.4 — Arc length and sector area.

Step 01 of 04

An arc length is a portion of the circumference; a sector area is a pie-slice of the disk. Both scale by the FRACTION of the circle the central angle covers.

fraction = \frac{central angle}{360°}

Step 02 of 04

Arc length = fraction × full circumference:

L = \frac{θ}{360°} \cdot 2 π r

Worked example: circle of radius 9 with a $80°$ arc.

L = \frac{80}{360} \cdot 2 π (9) = \frac{2}{9} \cdot 18 π = 4 π

Step 03 of 04

Sector area = fraction × full area:

A = \frac{θ}{360°} \cdot π r^{2}

Same circle (radius 9, $80°$ sector):

A = \frac{80}{360} \cdot π (9)^{2} = \frac{2}{9} \cdot 81 π = 18 π

Step 04 of 04

Common-fraction shortcuts.

90° (quarter)	→arc = $\frac{1}{4} \cdot 2 π r$ , area = $\frac{1}{4} \cdot π r^{2}$
180° (half)	→arc = $π r$ , area = $\frac{1}{2} π r^{2}$
60° (sixth)	→arc = $\frac{1}{6} \cdot 2 π r = \frac{π r}{3}$
120° (third)	→arc = $\frac{2 π r}{3}$

Key insight

Both formulas have the same shape: fraction × full thing. Arc length uses the circumference; sector area uses the area. Identify the fraction first (

θ /360°

), multiply by

2 π r

π r^{2}

as appropriate.

GE.G.5 — Equation of a circle.

Step 01 of 04

A circle with center $(h, k)$ and radius $r$ has the standard equation:

(x - h)^{2} + (y - k)^{2} = r^{2}

Reads as "distance from $(x, y)$ to $(h, k)$ equals $r$ " — squared on both sides to remove the radical. This is just the distance formula in disguise.

Step 02 of 04

Read the center and radius off the equation.

$(x - 3)^{2} + (y + 2)^{2} = 25$	→center $(3, - 2)$ , radius $5$
$x^{2} + y^{2} = 16$	→center $(0, 0)$ , radius $4$
$(x + 5)^{2} + y^{2} = 9$	→center $(- 5, 0)$ , radius $3$

The signs flip when reading h and k (just like vertex form for parabolas): $x + 5 = x - (- 5)$ , so $h = - 5$ . Radius is $r^{2}$ , not $r^{2}$ itself.

Step 03 of 04

From general form to standard form — complete the square.

Given $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ :

(x^{2} - 6 x) + (y^{2} + 4 y) (x^{2} - 6 x + 9) + (y^{2} + 4 y + 4) (x - 3)^{2} + (y + 2)^{2} = 12 = 12 + 9 + 4 = 25

Center $(3, - 2)$ , radius $5$ . Same as the algebra-1 completing-the-square move, applied to both x and y separately.

Step 04 of 04

Write the equation given center and a point on the circle. Find $r$ using the distance formula, then plug in.

Example: center $(2, - 1)$ , passes through $(5, 3)$ .

r = (5 - 2)^{2} + (3 - (- 1))^{2} = 9 + 16 = 5

(x - 2)^{2} + (y + 1)^{2} = 25

Key insight

(x - h)^{2} + (y - k)^{2} = r^{2}

— center signs flip, radius is the SQUARE ROOT of the right side. From general form, complete the square in x and in y separately to recover standard form.

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