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← SAT Math: Extension Workshops levels

SAT Math: Extension Workshops / LEVEL 2 · DIFFICULTY 3/5

SAT: Circles in Geometry & Coordinates

Focused SAT extension practice.

3 stages · 24 practice problems · two 6-question assessment forms

Choose a node to explore its lesson.

  1. 1MINI QUEST SAT 2.1Circle Equations & Completing the SquareRead the lesson
  2. 2MINI QUEST SAT 2.2Circle Angles, Chords & TangentsRead the lesson
  3. 3MINI QUEST SAT 2.3Arcs, Sectors & RadiansRead the lesson
  4. LEVEL CHECKCastle challenge

    6 questions across this level.

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STAGE SAT 2.1 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Circle Equations & Completing the Square

Useful preparation: Right-Triangle Distance

The idea to remember

In (x−h)2+(y−k)2=r2\displaystyle \left(x - h\right)^{2} + \left(y - k\right)^{2} = r^{2}(x−h)2+(y−k)2=r2, the center is (h,k) and radius r>0\displaystyle r > 0r>0. Complete both squares for general form.

Remember with 2 flashcards →

Goal: Read or construct a circle equation, complete squares, and check point membership using distance.

Before you begin: Pythagorean theorem, equations and angle measures.

Move the center. Keep every radius equal.

Parent equation: x2+y2=1\displaystyle x^{2} + y^{2} = 1x2+y2=1. This exploration is separate from the question’s values.

(1, 1)xy

(x−1)2+(y−1)2=4\displaystyle \left(x - 1\right)^{2} + \left(y - 1\right)^{2} = 4(x−1)2+(y−1)2=4. The right side is radius squared, so radius = 4=2\displaystyle \sqrt{4} = 24​=2.

Understand the idea

Parent circle: x2+y2=1\displaystyle x^{2} + y^{2} = 1x2+y2=1. Translating its center and scaling its radius gives (x−h)2+(y−k)2=r2\displaystyle \left(x - h\right)^{2} + \left(y - k\right)^{2} = r^{2}(x−h)2+(y−k)2=r2. The center is (h, k) and r>0\displaystyle r > 0r>0 is the radius.

Choose and justify a method

For x2\displaystyle x^{2}x2 + Dx, add (D2)2\displaystyle \left(\frac{D}{2}\right)^{2}(2D​)2 to form (x+D2)2\displaystyle \left(x + \frac{D}{2}\right)^{2}(x+2D​)2. Add the same quantity on the other side; do this separately for x and y. The right side after completing squares equals r2\displaystyle r^{2}r2, not r.

Check the conditions

Substitute a point into the left side to compute squared distance from the center. Equality with r2\displaystyle r^{2}r2 means on the circle; smaller means inside; larger means outside. A negative squared radius gives no real circle.

WORKED EXAMPLE 1

A circle has equation (x−4)2+(y+3)2=36\displaystyle \left(x - 4\right)^{2} + \left(y + 3\right)^{2} = 36(x−4)2+(y+3)2=36. What is its radius?

  1. Compare with the standard circle equation.
  2. The right side is radius squared.
  3. r2=36\displaystyle r^{2} = 36r2=36.
  4. The positive radius is 6.

6

WORKED EXAMPLE 2

A circular coverage region has center (4, 5) and passes through (7, 9). In its standard equation, what is the value of r2\displaystyle r^{2}r2?

  1. Subtract the center coordinates from the point coordinates.
  2. Use squared distance: horizontal change squared plus vertical change squared.
  3. The changes are 3 and 4.
  4. r2=32+42=25\displaystyle r^{2} = 3^{2} + 4^{2} = 25r2=32+42=25; this remains true after translating the center.

25

Common pitfalls

Possible mix-up: The radius is the right side of the standard equation.

Take the positive square root of the right side to obtain the radius.

Possible mix-up: The center has the same signs as the displayed constants.

Read (x−h)\displaystyle \left(x - h\right)(x−h) and (y−k)\displaystyle \left(y - k\right)(y−k); the center is (h, k).

Explain it to yourself

Explain why your method works in circle equations & completing the square. State one condition it needs and describe what can go wrong without that condition.

Discuss together

This is targeted topic practice in circle equations & completing the square. Ask for the reasoning and conditions, then compare independent solves, hint use and application attempts. Automatic checks assess the requested answer or proof step, not a complete written proof or university course completion.

Preview the eight practice prompts
  1. Problem 1

    A circle has equation (x−2)2+(y+1)2=16\displaystyle \left(x - 2\right)^{2} + \left(y + 1\right)^{2} = 16(x−2)2+(y+1)2=16. What is its radius?

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  2. Problem 2

    For the circle x2+y2−4 x−6 y=23\displaystyle x^{2} + y^{2} - 4\, x - 6\, y = 23x2+y2−4x−6y=23. What is the radius?

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  3. Problem 3

    A circle has diameter endpoints (0, 2) and (6, 2). Find the x-coordinate of its center.

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  4. Problem 4

    The circle has center (1, 2) and passes through (4, 6). In its standard equation, what is the value of r2\displaystyle r^{2}r2?

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  5. Problem 5New context

    A circular garden has equation (x−3)2+(y+2)2=25\displaystyle \left(x - 3\right)^{2} + \left(y + 2\right)^{2} = 25(x−3)2+(y+2)2=25. What is its radius?

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  6. Problem 6New context

    A circular sensor range is described by x2+y2−6 x−8 y=24\displaystyle x^{2} + y^{2} - 6\, x - 8\, y = 24x2+y2−6x−8y=24. What is the radius?

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  7. Problem 7New context

    A circular park has diameter endpoints (1, 2) and (9, 2). Find the x-coordinate of its center.

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  8. Problem 8New context

    A circular coverage region has center (2, 3) and passes through (5, 7). In its standard equation, what is the value of r2\displaystyle r^{2}r2?

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Open stage SAT 2.1 in the student workspace →

STAGE SAT 2.2 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Circle Angles, Chords & Tangents

Useful preparation: Circle Equations & Completing the Square

The idea to remember

An inscribed angle is half its intercepted arc's central angle. A tangent is perpendicular to the radius at contact.

Remember with 2 flashcards →

Goal: Select the correct circle theorem from the position of an angle, chord or tangent.

Before you begin: Pythagorean theorem, equations and angle measures.

Understand the idea

A central angle has its vertex at the center. An inscribed angle has its vertex on the circle and measures half its intercepted arc (the arc not containing its vertex). Equal intercepted arcs give equal inscribed angles.

Choose and justify a method

A tangent meets the circle at one point and is perpendicular to the radius there. A radius perpendicular to a chord bisects the chord. Draw that right triangle before using the Pythagorean theorem.

Check the conditions

For intersecting chords inside a circle, products of segment lengths agree: PA×PB=PC×PD\displaystyle PA \times PB = PC \times PDPA×PB=PC×PD. Tangent segments from the same external point are equal. Check that the diagram meets a theorem’s hypotheses before applying it.

WORKED EXAMPLE 1

An inscribed angle intercepts an arc measuring 88∘\displaystyle 88^{\circ}88∘. What is the angle in degrees?

  1. Locate the vertex on the circle.
  2. An inscribed angle is half its intercepted arc.
  3. Angle = 882\displaystyle \frac{88}{2}288​.
  4. It measures 44∘\displaystyle 44^{\circ}44∘.

44

WORKED EXAMPLE 2

A circular panel has radius 20 and a chord at perpendicular distance 12 from the center. Find the chord length.

  1. The perpendicular from the center bisects the chord.
  2. First find the half-chord with the Pythagorean theorem.
  3. Half-chord² = 202−122=256\displaystyle 20^{2} - 12^{2} = 256202−122=256.
  4. Half-chord = 16, so the whole chord is 32.

32

Common pitfalls

Possible mix-up: An inscribed angle equals its intercepted arc.

An inscribed angle is half the arc it intercepts.

Possible mix-up: Any chord is perpendicular to any radius.

Only the radius through the chord midpoint is perpendicular to that chord.

Explain it to yourself

Explain why your method works in circle angles, chords & tangents. State one condition it needs and describe what can go wrong without that condition.

Discuss together

This is targeted topic practice in circle angles, chords & tangents. Ask for the reasoning and conditions, then compare independent solves, hint use and application attempts. Automatic checks assess the requested answer or proof step, not a complete written proof or university course completion.

Preview the eight practice prompts
  1. Problem 1

    An inscribed angle intercepts an arc measuring 80∘\displaystyle 80^{\circ}80∘. What is the angle in degrees?

    Practise this exact problem →
  2. Problem 2

    A radius meets a tangent at T. If an adjacent angle within that right angle is 20∘\displaystyle 20^{\circ}20∘, what is the remaining angle in degrees?

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  3. Problem 3

    Two chords intersect. One chord has segment lengths 2 and 12; the other has segment lengths 4 and x. Find x.

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  4. Problem 4

    A circle has radius 5 and a chord at perpendicular distance 3 from the center. Find the chord length.

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  5. Problem 5New context

    An angle at the edge of a circular window intercepts an arc measuring 84∘\displaystyle 84^{\circ}84∘. What is the angle in degrees?

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  6. Problem 6New context

    A spoke of a circular wheel meets a tangent at T. If an adjacent angle within that right angle is 21∘\displaystyle 21^{\circ}21∘, what is the remaining angle in degrees?

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  7. Problem 7New context

    Two braces joining points on a circular frame intersect. One chord has segment lengths 3 and 15; the other has segment lengths 5 and x. Find x.

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  8. Problem 8New context

    A circular panel has radius 10 and a chord at perpendicular distance 6 from the center. Find the chord length.

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Open stage SAT 2.2 in the student workspace →

STAGE SAT 2.3 · 8 PRACTICE PROBLEMS · 4 NEW-CONTEXT APPLICATIONS

Arcs, Sectors & Radians

Useful preparation: Circle Angles, Chords & Tangents

The idea to remember

With θ\displaystyle \thetaθ in radians, arc length is r θ\displaystyle r\, \thetarθ and sector area is r2 θ2\displaystyle \frac{r^{2}\, \theta}{2}2r2θ​.

Remember with 2 flashcards →

Goal: Convert angle units and connect arc length, sector area and unit-circle coordinates.

Before you begin: Pythagorean theorem, equations and angle measures.

Understand the idea

One full turn is 360∘\displaystyle 360^{\circ}360∘ or 2 π\displaystyle 2\, \pi2π radians. Multiply degrees by π180\displaystyle \frac{\pi}{180}180π​ to convert to radians. A radian measures arc length divided by radius.

Choose and justify a method

With θ\displaystyle \thetaθ in radians, arc length s=r θ\displaystyle s = r\, \thetas=rθ and sector area A=r2 θ2\displaystyle A = \frac{r^{2}\, \theta}{2}A=2r2θ​. With degrees, use the fraction θ360\displaystyle \frac{\theta}{360}360θ​ of the circumference or area. Do not put a degree value directly into the radian formula.

Check the conditions

On the unit circle, the point at angle θ\displaystyle \thetaθ is (cos⁡ θ,sin⁡ θ)\displaystyle \left(\operatorname{cos}\, \theta, \operatorname{sin}\, \theta\right)(cosθ,sinθ). The signs come from its quadrant. Scaling to radius r multiplies both coordinates by r; translating adds the center coordinates.

WORKED EXAMPLE 1

A sector has radius 5 and central angle 60∘\displaystyle 60^{\circ}60∘. Its arc length is a π\displaystyle a\, \piaπ. Find a.

  1. 60∘\displaystyle 60^{\circ}60∘ is one sixth of a full circle.
  2. Multiply 2 π r\displaystyle 2\, \pi\, r2πr by 16\displaystyle \frac{1}{6}61​.
  3. Arc = 2 π×56\displaystyle \frac{2\, \pi \times 5}{6}62π×5​.
  4. The coefficient of π\displaystyle \piπ is 53\displaystyle \frac{5}{3}35​.

53\displaystyle \frac{5}{3}35​

WORKED EXAMPLE 2

A marker on a wheel lies on a circle of radius 5, centered at the origin, at angle π\displaystyle \piπ radians from the positive x-axis. What is its x-coordinate?

  1. π\displaystyle \piπ radians is a half turn.
  2. The point lies on the negative x-axis.
  3. x=5 cos⁡ π\displaystyle x = 5\, \operatorname{cos}\, \pix=5cosπ.
  4. Since cos⁡ π=−1\displaystyle \operatorname{cos}\, \pi = -1cosπ=−1, x=−5\displaystyle x = -5x=−5.

-5

Common pitfalls

Possible mix-up: s=r θ\displaystyle s = r\, \thetas=rθ works unchanged for degrees.

Convert θ\displaystyle \thetaθ to radians first, or use θ360\displaystyle \frac{\theta}{360}360θ​ of the circumference.

Possible mix-up: A larger radius changes the angle of a fixed fraction of a turn.

The fraction of a full turn fixes the angle; radius changes lengths and areas.

Explain it to yourself

Explain why your method works in arcs, sectors & radians. State one condition it needs and describe what can go wrong without that condition.

Discuss together

This is targeted topic practice in arcs, sectors & radians. Ask for the reasoning and conditions, then compare independent solves, hint use and application attempts. Automatic checks assess the requested answer or proof step, not a complete written proof or university course completion.

Preview the eight practice prompts
  1. Problem 1

    A sector has radius 3 and central angle 60∘\displaystyle 60^{\circ}60∘. Its arc length is a π\displaystyle a\, \piaπ. Find a.

    Practise this exact problem →
  2. Problem 2

    A sector has radius 2 and angle π2\displaystyle \frac{\pi}{2}2π​ radians. Its area is b π\displaystyle b\, \pibπ. Find b.

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  3. Problem 3

    An angle measures 60∘\displaystyle 60^{\circ}60∘. Write the angle as a π\displaystyle a\, \piaπ radians. Find a.

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  4. Problem 4

    A point lies on a circle of radius 2, centered at the origin, at angle π\displaystyle \piπ radians from the positive x-axis. What is its x-coordinate?

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  5. Problem 5New context

    A circular fan sector has radius 4 and central angle 60∘\displaystyle 60^{\circ}60∘. Its arc length is a π\displaystyle a\, \piaπ. Find a.

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  6. Problem 6New context

    A round-table sector has radius 3 and angle π2\displaystyle \frac{\pi}{2}2π​ radians. Its area is b π\displaystyle b\, \pibπ. Find b.

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  7. Problem 7New context

    A wheel turns through 90∘\displaystyle 90^{\circ}90∘. Write the angle as a π\displaystyle a\, \piaπ radians. Find a.

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  8. Problem 8New context

    A marker on a wheel lies on a circle of radius 3, centered at the origin, at angle π\displaystyle \piπ radians from the positive x-axis. What is its x-coordinate?

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Open stage SAT 2.3 in the student workspace →
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∑ Numerive

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HomeTry a questionMy questsLearning SearchHomeworkUnderstand & rememberCurriculumFlashcardsUniversity

Exam preparation

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SAT: Circles in Geometry & Coordinates: lessons and practice | Numerive Math