Digital SAT Maths -Unit 4 - 4.7 Circles- Study Notes- New Syllabus
Digital SAT Maths -Unit 4 – 4.7 Circles- Study Notes- New syllabus
Digital SAT Maths -Unit 4 – 4.7 Circles- Study Notes – per latest Syllabus.
Key Concepts:
Arc length
Sector area
Different Types of Equations of a Circle
Chords & tangents Properties
Arc Length
An arc is a portion of the circumference of a circle. Arc length tells us the distance along the curved edge of the circle.
The full circumference of a circle is:
\( C = 2\pi r \)
If the central angle does not cover the full circle, we take a fraction of the circumference.
Arc Length Formula (degrees)
\( \text{Arc Length} = \dfrac{\theta}{360^\circ}\,(2\pi r) \)
where:
- \( \theta \) = central angle
- \( r \) = radius
Arc Length Formula (radians)
\( \text{Arc Length} = r\theta \)
(This formula works only when \( \theta \) is in radians.)
DIGITAL SAT Tip
Most SAT problems use degrees, not radians, so be careful which formula to use.
Example 1:
Find the arc length of a circle with radius 7 cm and central angle \( 90^\circ \).
▶️ Answer/Explanation
\( L = \dfrac{90}{360}(2\pi\cdot7) \)
\( L = \dfrac{1}{4}(14\pi) \)
\( L = \dfrac{14\pi}{4} = \dfrac{7\pi}{2} \)
Answer: \( \dfrac{7\pi}{2} \) cm
Example 2:
A circular track has radius 20 m. A runner covers an angle of \( 180^\circ \). How far did the runner travel?
▶️ Answer/Explanation
\( L = \dfrac{180}{360}(2\pi\cdot20) \)
\( L = \dfrac{1}{2}(40\pi) \)
\( L = 20\pi \)
Answer: \( 20\pi \) m
Example 3 (Find Angle):
A circle has radius 10 cm and an arc length of \( 5\pi \) cm. Find the central angle in degrees.
▶️ Answer/Explanation
\( 5\pi = \dfrac{\theta}{360}(2\pi\cdot10) \)
\( 5\pi = \dfrac{\theta}{360}(20\pi) \)
Cancel \( \pi \):
\( 5 = \dfrac{20\theta}{360} \)
\( 5 = \dfrac{\theta}{18} \)
\( \theta = 90^\circ \)
Answer: \( 90^\circ \)
Sector Area
A sector is a region of a circle bounded by two radii and the arc between them. It looks like a “slice of pizza”.
The area of a full circle is:
\( A = \pi r^2 \)
A sector is just a fraction of the full circle.
Sector Area Formula (degrees)
\( \text{Sector Area} = \dfrac{\theta}{360^\circ}\,(\pi r^2) \)
where:
- \( \theta \) = central angle
- \( r \) = radius
Sector Area (radians)
\( \text{Sector Area} = \dfrac{1}{2}r^2\theta \)
(only if \( \theta \) is in radians)
Connection With Arc Length
If arc length is \( L \):
\( \text{Sector Area} = \dfrac{1}{2}rL \)
DIGITAL SAT Tip
Be careful not to confuse arc length with sector area. Arc length is distance along the curve, sector area is the region inside.
Example 1:
Find the area of a sector with radius 6 cm and central angle \( 60^\circ \).
▶️ Answer/Explanation
\( A = \dfrac{60}{360}(\pi\cdot6^2) \)
\( A = \dfrac{1}{6}(36\pi) \)
\( A = 6\pi \)
Answer: \( 6\pi \text{ cm}^2 \)
Example 2:
A pizza has radius 10 cm. A slice forms a \( 90^\circ \) sector. Find the area of the slice.
▶️ Answer/Explanation
\( A = \dfrac{90}{360}(\pi\cdot10^2) \)
\( A = \dfrac{1}{4}(100\pi) \)
\( A = 25\pi \)
Answer: \( 25\pi \text{ cm}^2 \)
Example 3 (Using Arc Length):
A circle has radius 8 cm and arc length \( 4\pi \) cm. Find the sector area.
▶️ Answer/Explanation
\( A=\dfrac{1}{2}rL \)
\( A=\dfrac{1}{2}(8)(4\pi) \)
\( A=16\pi \)
Answer: \( 16\pi \text{ cm}^2 \)
Different Types of Equations of a Circle
On the DIGITAL SAT, circles often appear in coordinate geometry. You must recognize the equation of a circle and identify its center and radius.
Standard Form of a Circle
\( (x – h)^2 + (y – k)^2 = r^2 \)
where:
- \( (h,k) \) = center of the circle
- \( r \) = radius
Important Note
The signs are opposite inside the brackets:
\( (x – 3)^2 \) → center \( x = 3 \)
\( (y + 2)^2 \) → center \( y = -2 \)
Circle at the Origin
If the center is \( (0,0) \):
\( x^2 + y^2 = r^2 \)
Expanded (General) Form
Sometimes the equation is given as:
\( x^2 + y^2 + Dx + Ey + F = 0 \)
You must convert it to standard form by completing the square.
DIGITAL SAT Tip
The most common questions:
- Find the center
- Find the radius
- Check whether a point lies on the circle
Example 1 (Identify Center and Radius):
Find the center and radius of
\( (x-4)^2 + (y+1)^2 = 25 \)
▶️ Answer/Explanation
Center:
\( (4,-1) \)
Radius:
\( r=\sqrt{25}=5 \)
Answer: center \( (4,-1) \), radius 5
Example 2 (Circle at Origin):
Does the point \( (3,4) \) lie on the circle
\( x^2 + y^2 = 25 \) ?
▶️ Answer/Explanation
\( 3^2 + 4^2 = 9 + 16 = 25 \)
It satisfies the equation.
Answer: Yes, the point lies on the circle.
Example 3 (Convert to Standard Form):
Find the center and radius of
\( x^2 + y^2 – 6x + 8y – 11 = 0 \)
▶️ Answer/Explanation
Group terms:
\( (x^2 – 6x) + (y^2 + 8y) = 11 \)
Complete squares:
\( x^2 – 6x + 9 + y^2 + 8y + 16 = 11 + 9 + 16 \)
\( (x-3)^2 + (y+4)^2 = 36 \)
Center \( (3,-4) \), radius \( 6 \).
Chords & Tangents Properties
The DIGITAL SAT often tests geometric properties of circles instead of heavy algebra. You must recognize relationships involving chords and tangents.
Chord
A chord is a line segment whose endpoints lie on the circle.
Diameter is the longest chord and passes through the center.
Tangent
A tangent is a line that touches the circle at exactly one point.
Key Properties
- A radius to a tangent is perpendicular to the tangent
- Equal chords are equally distant from the center
- The perpendicular from the center to a chord bisects the chord
Tangent–Radius Theorem
Radius ⟂ Tangent at the point of contact
This means a right angle (90°) is always formed.
Tangent Length Property
Tangents drawn from the same external point are equal:
\( PA = PB \)
DIGITAL SAT Tip
Whenever a tangent appears, look for a hidden right triangle.
Example 1 (Radius and Tangent):
A radius meets a tangent at point \( T \). Find the angle between them.
▶️ Answer/Explanation
A radius is perpendicular to a tangent.
Answer: \( 90^\circ \)
Example 2 (Equal Tangents):
From an external point \( P \), two tangents \( PA \) and \( PB \) are drawn to a circle. If \( PA = 12 \), find \( PB \).
▶️ Answer/Explanation
Tangents from the same external point are equal.
Answer: \( PB = 12 \)
Example 3 (Chord Property):
A perpendicular from the center of a circle meets a chord of length 10. Find the length of each half.
▶️ Answer/Explanation
The perpendicular bisects the chord.
\( \dfrac{10}{2}=5 \)
Answer: each half = 5