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Radians in circle geometry problems

Solve radian geometry problems involving tangents, inscribed circles and overlapping circles. Build consistent diagrams and separate curved areas from their boundaries.

Before you startArc length, sector and segment area, right-triangle trigonometry and circle theorems

01 / Start with the geometry

An area formula is useful only after the shape is understood.

Mark centres, radii, tangents and shared chords before calculating. Decide which arcs enclose the requested region, and whether a straight line lies on the outside or only divides the interior.

Area: split the region into known pieces.
Perimeter: trace only the outside boundary.

The model keeps both circles geometrically consistent as you change their separation. Shade either segment to see how the lens is assembled.

Two circles, one shared chordExplore

α ≈1.854590; β ≈1.854590 radians.

Circle A segment ≈11.182380; circle B segment ≈11.182380; lens ≈22.364761 square units.

Lens boundary ≈18.545904 units.

The lens is two segments. Its boundary is two arcs; the shared chord is internal and is not part of the perimeter.

Circle B radiusDistance between centres 6Shade

Watch two segments form a lens

Pause, replay or seek freely. The notes explain the same idea and stay in view.

02 / Use circle facts before formulas

A short geometric argument can reveal the central angle.

A radius meets a tangent at a right angle.
Two tangents drawn from the same external point have equal lengths.
The perpendicular from a circle’s centre to a chord bisects that chord.
The central angle is twice an angle at the circumference standing on the same arc.
Angles in a triangle total π; angles around a point total 2π.

An inscribed angle is π/5Worked example

Central angle on the same arc =2π/5

Check that both angles stand on the same chosen arc.

For radius 10, arc length =10(2π/5)=4π

The opposite major arc would use 8π/5.

03 / Find a tangent-bounded region

Join the centre to the external point to make two right triangles.

From an external point P, draw tangents PA and PB to a circle with centre O and radius r. Let ∠AOP =α, with 0<α<π/2. The two right triangles are congruent, so ∠AOB =2α.

PA =PB =r tan α
OP =r/cos α
Area between the tangents and minor arc AB =r²(tan α −α)
Boundary length =2r tan α +2rα

The kite OAPB consists of two right triangles, each of area ½r²tan α. Subtract the sector with angle 2α. The region’s boundary contains both tangent lengths and the small arc; the radii are internal construction lines.

r =6 and α =π/6Worked example

Tangent length =6/√3 =2√3

The central angle is π/3.

Area =12√3 −6π

Kite area 12√3 minus sector area 6π.

Boundary =4√3 +2π

Two tangents plus the minor arc.

04 / Fit a circle inside a sector

The small centre lies on the sector’s angle bisector.

A circle of radius ρ touches both straight sides of a sector of radius R and angle θ, where 0<θ<π, and touches its outer arc internally. If its centre is C and the sector centre is O, then OC =R−ρ.

Derive the radiusWorked example

ρ =(R−ρ)sin(θ/2)

Drop a perpendicular from C to either straight side; this perpendicular is a small radius.

ρ[1+sin(θ/2)] =R sin(θ/2)

Collect the terms containing ρ.

ρ =R sin(θ/2)/[1+sin(θ/2)]

Internal tangency supplies OC =R−ρ; do not use R+ρ.

For R =12 and θ =π/3, ρ =12(1/2)/(3/2)=4. The sector area is 24π and the small-circle area is 16π, so the part outside the small circle has area 8π.

05 / Subtract a sector from a triangle

Check that the sector really lies inside the triangle.

In triangle OAB, OA =OB =10 and ∠AOB =π/3. A sector centred at O with radius 4 uses the two rays OA and OB. Its arc lies inside the triangle because the distance from O to AB is 10cos(π/6)=5√3, greater than 4.

Remaining area and perimeterWorked example

Triangle area =½(10)(10)sin(π/3)=25√3

The triangle is equilateral, so AB =10.

Sector area =½(4²)(π/3)=8π/3

Subtract this from the triangle.

Remaining area =25√3 −8π/3

Square units.

Boundary =(10−4)+(10−4)+10+4π/3 =22+4π/3

The two removed radius lengths are replaced by the curved arc.

06 / Make the lens data agree

Both circles must produce exactly the same common chord.

Let the radii be a and b, the centre separation be d, and let x be the distance from centre A to the shared chord along AB. If the chord lies between the centres, its half-length h satisfies two right-triangle equations.

x²+h² =a²
(d−x)²+h² =b²
x =(d²+a²−b²)/(2d)
h =√(a²−x²)

Then α =2arccos(x/a), β =2arccos[(d−x)/b], and the shared chord is 2h. Equivalently, 2a sin(α/2)=2b sin(β/2). Independently choosing two radii and two angles usually fails this consistency condition.

The model uses crossing circles with the chord between their centres. Partial overlap requires |a−b|<d<a+b. For other placements the same coordinate equations still help, but an overlap may require a major segment; draw the regions before adding areas.

07 / Add the segments and trace the arcs

The shared chord separates the pieces but disappears from the boundary.

Lens area =½a²(α−sin α)+½b²(β−sin β)
Lens perimeter =aα+bβ

Here the common chord lies between the centres, as in the model.

Two radius 5 circles with centres 6 apartWorked example

x =3, h =4, common chord =8

The circles are equal, so the chord is halfway between the centres.

α =β =2arccos(3/5)

Either inverse cosine or inverse sine of 4/5 gives the half-angle.

Each segment =25arccos(3/5)−12

The triangle in each sector has base 8 and height 3.

Lens area =50arccos(3/5)−24

Approximately 22.36476 square units.

Lens perimeter =20arccos(3/5)

Approximately 18.54590 units; do not add the shared chord.

08 / Check feasibility and limiting behaviour

A numerical answer should fit the diagram.

If d>a+b, the circles do not meet. If d=a+b, they touch externally and the lens has zero area.
If d<|a−b|, one circle is completely inside the other; the overlap is the smaller disk.
If d=0 and a=b, the circles coincide; division by d is invalid even though the overlap is clear.
An overlap area cannot exceed either full-circle area.
As equal circles move apart, their common lens shrinks; close to external tangency both boundary arcs become short.

A tangent or a centre-to-chord distance cannot be negative. Check that an inverse sine/cosine input lies in [−1,1] before treating it as an angle.

09 / Keep parameters until the last step

Exact relationships often matter more than a decimal.

The incircle radius is one third of the sector radiusWorked example

ρ/R =1/3 =s/(1+s), where s=sin(θ/2)

Use the sector-incircle relationship.

1+s =3s, so s=1/2

The radius scale cancels.

A circle inscribed in a sector has radius R/3, where R is the sector radius. Find the sector angle θ, assuming 0<θ<π.

Hint

Let s=sin(θ/2) and solve 1/3=s/(1+s).

Worked solution

s=1/2; θ/2=π/6; θ=π/3.

11 / Recap

Let the geometry decide which formulas apply.

Use right angles at tangents and half-chord triangles.
An inscribed circle touches internally: the centre distance is R−ρ.
Check that a subtracted sector fits inside the triangle.
A consistent lens uses one common chord.
Add segment areas; add only the external boundary arcs.
Check overlap conditions, dimensions and exact angle ranges.

Section 1 of 11 · Start with the geometry