Study Circles in PSAT Math with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: State the chord-chord power theorem: chords intersect at E E E with segments A E , E B AE,EB A E , EB and C E , E D CE,ED CE , E D . Answer: A E ⋅ E B = C E ⋅ E D AE\cdot EB=CE\cdot ED A E ⋅ EB = CE ⋅ E D . Products of chord segments are equal at intersection.
Flashcard 2: Find the arc length when r = 6 r=6 r = 6 and θ = 60 ∘ \theta=60^\circ θ = 6 0 ∘ . Answer: 2 π 2\pi 2 π . s = 60 360 ⋅ 2 π ( 6 ) = 1 6 ⋅ 12 π = 2 π s=\frac{60}{360}\cdot 2\pi(6)=\frac{1}{6}\cdot 12\pi=2\pi s = 360 60 ⋅ 2 π ( 6 ) = 6 1 ⋅ 12 π = 2 π .
Flashcard 3: State the equation of a circle with center ( h , k ) (h,k) ( h , k ) and radius r r r in standard form. Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Standard form shows center coordinates and radius squared.
Flashcard 4: What is the relationship between diameter d d d and radius r r r in a circle? Answer: d = 2 r d=2r d = 2 r . Diameter is twice the radius.
Flashcard 5: Find the distance between the centers of two circles with centers ( 1 , 2 ) (1,2) ( 1 , 2 ) and ( 4 , 6 ) (4,6) ( 4 , 6 ) . Answer: 5 5 5 . Distance formula: ( 4 − 1 ) 2 + ( 6 − 2 ) 2 = 9 + 16 = 5 \sqrt{(4-1)^2+(6-2)^2}=\sqrt{9+16}=5 ( 4 − 1 ) 2 + ( 6 − 2 ) 2 = 9 + 16 = 5 .
Flashcard 6: What is the measure of an inscribed angle that intercepts an arc of measure m m m degrees? Answer: m 2 \frac{m}{2} 2 m . Inscribed angle theorem: half the intercepted arc.
Flashcard 7: State the area formula for a circle with radius r r r . Answer: A = π r 2 A=\pi r^2 A = π r 2 . Area equals π \pi π times radius squared.
Flashcard 8: Identify the radius of the circle ( x + 1 ) 2 + ( y − 4 ) 2 = 49 (x+1)^2+(y-4)^2=49 ( x + 1 ) 2 + ( y − 4 ) 2 = 49 . Answer: 7 7 7 . Radius equals the square root of the constant term.
Flashcard 9: Identify the center and radius of x 2 + y 2 − 8 x + 6 y = 0 x^2+y^2-8x+6y=0 x 2 + y 2 − 8 x + 6 y = 0 . Answer: Center ( 4 , − 3 ) (4,-3) ( 4 , − 3 ) , radius 5 5 5 . Complete the square: ( x − 4 ) 2 + ( y + 3 ) 2 = 25 (x-4)^2+(y+3)^2=25 ( x − 4 ) 2 + ( y + 3 ) 2 = 25 .
Flashcard 10: Find the arc length for r = 6 r=6 r = 6 and θ = 60 ∘ \theta=60^\circ θ = 6 0 ∘ . Answer: 2 π 2\pi 2 π . s = 60 360 ⋅ 2 π ( 6 ) = 1 6 ⋅ 12 π = 2 π s=\frac{60}{360}\cdot 2\pi(6)=\frac{1}{6}\cdot 12\pi=2\pi s = 360 60 ⋅ 2 π ( 6 ) = 6 1 ⋅ 12 π = 2 π .
Flashcard 11: What is the length of a semicircle arc with radius r r r (arc only, not including the diameter)? Answer: π r \pi r π r . Half the circumference gives the semicircle arc.
Flashcard 12: Find the measure of an inscribed angle that intercepts a 110 ∘ 110^\circ 11 0 ∘ arc. Answer: 55 ∘ 55^\circ 5 5 ∘ . Inscribed angle is half the intercepted arc: 110 ° 2 = 55 ° \frac{110°}{2}=55° 2 110° = 55° .
Flashcard 13: Find the radius if the center is ( 2 , 3 ) (2,3) ( 2 , 3 ) and a point on the circle is ( 2 , 11 ) (2,11) ( 2 , 11 ) . Answer: 8 8 8 . Distance from ( 2 , 3 ) (2,3) ( 2 , 3 ) to ( 2 , 11 ) (2,11) ( 2 , 11 ) is ∣ 11 − 3 ∣ = 8 |11-3|=8 ∣11 − 3∣ = 8 .
Flashcard 14: State the standard form equation of a circle with center ( h , k ) (h,k) ( h , k ) and radius r r r . Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Standard form shows center coordinates and radius squared.
Flashcard 15: Find the radius of the circle ( x + 5 ) 2 + ( y − 7 ) 2 = 49 (x+5)^2+(y-7)^2=49 ( x + 5 ) 2 + ( y − 7 ) 2 = 49 . Answer: 7 7 7 . r 2 = 49 r^2=49 r 2 = 49 , so r = 49 = 7 r=\sqrt{49}=7 r = 49 = 7 .
Flashcard 16: Use P T 2 = P A ⋅ P B PT^2=PA\cdot PB P T 2 = P A ⋅ PB : if P A = 4 PA=4 P A = 4 and P B = 16 PB=16 PB = 16 , find P T PT PT . Answer: 8 8 8 . P T 2 = 4 ⋅ 16 = 64 PT^2=4\cdot 16=64 P T 2 = 4 ⋅ 16 = 64 , so P T = 64 = 8 PT=\sqrt{64}=8 PT = 64 = 8 .
Flashcard 17: What is the diameter of a circle in terms of radius r r r ? Answer: d = 2 r d=2r d = 2 r . Diameter is twice the radius.
Flashcard 18: State the tangent-secant power theorem from external point P P P : tangent P T PT PT , secant meets circle at A A A and B B B . Answer: P T 2 = P A ⋅ P B PT^2=PA\cdot PB P T 2 = P A ⋅ PB . Tangent squared equals product of secant segments from P.
Flashcard 19: What is the circumference formula of a circle with radius r r r ? Answer: C = 2 π r C=2\pi r C = 2 π r . Circumference equals 2 π 2\pi 2 π times the radius.
Flashcard 20: State the standard form of a circle with center ( h , k ) (h,k) ( h , k ) and radius r r r . Answer: ( x − h ) 2 + ( y − k ) 2 = r 2 (x-h)^2+(y-k)^2=r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 . Standard form shows center and radius directly.
Flashcard 21: State the circumference formula for a circle with radius r r r . Answer: C = 2 π r C=2\pi r C = 2 π r . Circumference equals 2 π 2\pi 2 π times the radius.
Flashcard 22: What are the center and radius of ( x − 3 ) 2 + ( y + 2 ) 2 = 25 (x-3)^2+(y+2)^2=25 ( x − 3 ) 2 + ( y + 2 ) 2 = 25 ? Answer: Center ( 3 , − 2 ) (3,-2) ( 3 , − 2 ) , radius 5 5 5 . Read ( h , k ) (h,k) ( h , k ) from ( x − h ) 2 + ( y − k ) 2 (x-h)^2+(y-k)^2 ( x − h ) 2 + ( y − k ) 2 and r r r from r 2 = 25 r^2=25 r 2 = 25 .
Flashcard 23: What is the arc length formula for central angle θ \theta θ in degrees and radius r r r ? Answer: s = θ 360 ⋅ 2 π r s=\frac{\theta}{360}\cdot 2\pi r s = 360 θ ⋅ 2 π r . Arc length is the fraction of circumference.
Flashcard 24: Identify the center and radius from ( x − 3 ) 2 + ( y + 4 ) 2 = 25 (x-3)^2+(y+4)^2=25 ( x − 3 ) 2 + ( y + 4 ) 2 = 25 . Answer: Center ( 3 , − 4 ) (3,-4) ( 3 , − 4 ) , radius 5 5 5 . Read ( h , k ) (h,k) ( h , k ) from ( x − h ) 2 + ( y − k ) 2 (x-h)^2+(y-k)^2 ( x − h ) 2 + ( y − k ) 2 and r = 25 = 5 r=\sqrt{25}=5 r = 25 = 5 .
Flashcard 25: Find the center of the circle ( x + 5 ) 2 + ( y − 7 ) 2 = 49 (x+5)^2+(y-7)^2=49 ( x + 5 ) 2 + ( y − 7 ) 2 = 49 . Answer: ( − 5 , 7 ) (-5,7) ( − 5 , 7 ) . Center is ( h , k ) (h,k) ( h , k ) where equation is ( x − h ) 2 + ( y − k ) 2 (x-h)^2+(y-k)^2 ( x − h ) 2 + ( y − k ) 2 .
Flashcard 26: State the relationship between diameter d d d and radius r r r . Answer: d = 2 r d=2r d = 2 r . Diameter is twice the radius.
Flashcard 27: What is the area formula of a circle with radius r r r ? Answer: A = π r 2 A=\pi r^2 A = π r 2 . Area equals π \pi π times radius squared.
Flashcard 28: What is the radius of a circle in terms of diameter d d d ? Answer: r = d 2 r=\frac{d}{2} r = 2 d . Radius is half the diameter.
Flashcard 29: What is the distance formula used to find a radius from center ( h , k ) (h,k) ( h , k ) to point ( x , y ) (x,y) ( x , y ) ? Answer: r = ( x − h ) 2 + ( y − k ) 2 r=\sqrt{(x-h)^2+(y-k)^2} r = ( x − h ) 2 + ( y − k ) 2 . Distance formula gives radius from center to any point.
Flashcard 30: What is the measure of a central angle that intercepts an arc of measure m m m degrees? Answer: m m m . Central angle equals its intercepted arc measure.
Flashcard 31: Find the area of a sector for r = 3 r=3 r = 3 and θ = 120 ∘ \theta=120^\circ θ = 12 0 ∘ . Answer: 3 π 3\pi 3 π . A = 120 360 ⋅ π ( 3 ) 2 = 1 3 ⋅ 9 π = 3 π A=\frac{120}{360}\cdot \pi(3)^2=\frac{1}{3}\cdot 9\pi=3\pi A = 360 120 ⋅ π ( 3 ) 2 = 3 1 ⋅ 9 π = 3 π .
Flashcard 32: State the sector area formula for central angle θ \theta θ (in degrees) and radius r r r . Answer: A = θ 360 ⋅ π r 2 A=\frac{\theta}{360}\cdot \pi r^2 A = 360 θ ⋅ π r 2 . Fraction of full area based on angle ratio.
Flashcard 33: State the equation of a circle centered at the origin with radius r r r . Answer: x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 . Center at origin means ( h , k ) = ( 0 , 0 ) (h,k)=(0,0) ( h , k ) = ( 0 , 0 ) .
Flashcard 34: State the arc length formula for central angle θ \theta θ in degrees and radius r r r . Answer: s = θ 360 ⋅ 2 π r s=\frac{\theta}{360}\cdot 2\pi r s = 360 θ ⋅ 2 π r . Arc length is the fraction of circumference for given angle.
Flashcard 35: Use A E ⋅ E B = C E ⋅ E D AE\cdot EB=CE\cdot ED A E ⋅ EB = CE ⋅ E D : if A E = 3 AE=3 A E = 3 , E B = 12 EB=12 EB = 12 , and C E = 6 CE=6 CE = 6 , find E D ED E D . Answer: 6 6 6 . 3 ⋅ 12 = 6 ⋅ E D 3\cdot 12=6\cdot ED 3 ⋅ 12 = 6 ⋅ E D , so E D = 36 6 = 6 ED=\frac{36}{6}=6 E D = 6 36 = 6 .
Flashcard 36: Find the arc length of a 90 ∘ 90^\circ 9 0 ∘ sector in a circle of radius 8 8 8 . Answer: 4 π 4\pi 4 π . Use s = 90 360 ⋅ 2 π ( 8 ) = 1 4 ⋅ 16 π s=\frac{90}{360}\cdot 2\pi(8)=\frac{1}{4}\cdot 16\pi s = 360 90 ⋅ 2 π ( 8 ) = 4 1 ⋅ 16 π .
Flashcard 37: Which statement is always true about a radius drawn to a point of tangency? Answer: It is perpendicular to the tangent line. Radius meets tangent at 90 ° 90° 90° angle.
Flashcard 38: Identify the center and radius of ( x − 3 ) 2 + ( y + 2 ) 2 = 25 (x-3)^2+(y+2)^2=25 ( x − 3 ) 2 + ( y + 2 ) 2 = 25 . Answer: Center ( 3 , − 2 ) (3,-2) ( 3 , − 2 ) , radius 5 5 5 . Read ( h , k ) (h,k) ( h , k ) from ( x − h ) 2 + ( y − k ) 2 (x-h)^2+(y-k)^2 ( x − h ) 2 + ( y − k ) 2 ; r 2 = 25 r^2=25 r 2 = 25 so r = 5 r=5 r = 5 .
Flashcard 39: State the circumference of a circle with diameter d d d . Answer: C = π d C=\pi d C = π d . Circumference equals π \pi π times diameter.
Flashcard 40: Find the area of a circle with diameter 10 10 10 . Answer: 25 π 25\pi 25 π . Use A = π r 2 A=\pi r^2 A = π r 2 with r = 5 r=5 r = 5 (half of diameter).
Flashcard 41: State the relationship between a tangent and the radius at the point of tangency. Answer: Tangent is perpendicular to the radius. Forms a 90 ° 90° 90° angle at the point of tangency.
Flashcard 42: Find the area of a sector when r = 10 r=10 r = 10 and θ = 90 ∘ \theta=90^\circ θ = 9 0 ∘ . Answer: 25 π 25\pi 25 π . A = 90 360 ⋅ π ( 10 ) 2 = 1 4 ⋅ 100 π = 25 π A=\frac{90}{360}\cdot \pi(10)^2=\frac{1}{4}\cdot 100\pi=25\pi A = 360 90 ⋅ π ( 10 ) 2 = 4 1 ⋅ 100 π = 25 π .
Flashcard 43: What is the sector area formula for central angle θ \theta θ in degrees and radius r r r ? Answer: A = θ 360 ⋅ π r 2 A=\frac{\theta}{360}\cdot \pi r^2 A = 360 θ ⋅ π r 2 . Sector area is the fraction of circle area.
Flashcard 44: State the tangent property relating a radius to a tangent line at the point of tangency. Answer: A radius is perpendicular to the tangent: r ⊥ tangent r\perp \text{tangent} r ⊥ tangent . Radius meets tangent at 90 ° 90° 90° angle.
Flashcard 45: Identify the radius of ( x + 1 ) 2 + ( y − 2 ) 2 = 49 4 (x+1)^2+(y-2)^2=\frac{49}{4} ( x + 1 ) 2 + ( y − 2 ) 2 = 4 49 . Answer: 7 2 \frac{7}{2} 2 7 . r 2 = 49 4 r^2=\frac{49}{4} r 2 = 4 49 , so r = 7 2 r=\frac{7}{2} r = 2 7 .
Flashcard 46: State the circle equation with center ( 2 , − 1 ) (2,-1) ( 2 , − 1 ) and radius 4 4 4 . Answer: ( x − 2 ) 2 + ( y + 1 ) 2 = 16 (x-2)^2+(y+1)^2=16 ( x − 2 ) 2 + ( y + 1 ) 2 = 16 . Substitute center ( h , k ) = ( 2 , − 1 ) (h,k)=(2,-1) ( h , k ) = ( 2 , − 1 ) and r 2 = 16 r^2=16 r 2 = 16 .
Flashcard 47: What is the length of a semicircle arc (not including the diameter) with radius r r r ? Answer: π r \pi r π r . Half the circumference, excluding the diameter.
Flashcard 48: State the arc length formula for central angle θ \theta θ (in degrees) and radius r r r . Answer: s = θ 360 ⋅ 2 π r s=\frac{\theta}{360}\cdot 2\pi r s = 360 θ ⋅ 2 π r . Fraction of full circumference based on angle ratio.
Flashcard 49: Find the circumference of a circle with diameter 12 12 12 . Answer: 12 π 12\pi 12 π . Use C = 2 π r C=2\pi r C = 2 π r with r = 6 r=6 r = 6 (half of diameter).
Flashcard 50: Find the area of a 60 ∘ 60^\circ 6 0 ∘ sector of a circle with radius 6 6 6 . Answer: 6 π 6\pi 6 π . Use A = 60 360 ⋅ π ( 6 ) 2 = 1 6 ⋅ 36 π A=\frac{60}{360}\cdot\pi(6)^2=\frac{1}{6}\cdot 36\pi A = 360 60 ⋅ π ( 6 ) 2 = 6 1 ⋅ 36 π .
Flashcard 51: State the formula for the circumference of a circle with radius r r r . Answer: C = 2 π r C=2\pi r C = 2 π r . Circumference equals 2 π 2\pi 2 π times the radius.
Flashcard 52: State the formula for the area of a circle with radius r r r . Answer: A = π r 2 A=\pi r^2 A = π r 2 . Area equals π \pi π times radius squared.
Flashcard 53: State the sector area formula for central angle θ \theta θ in degrees and radius r r r . Answer: A = θ 360 ⋅ π r 2 A=\frac{\theta}{360}\cdot \pi r^2 A = 360 θ ⋅ π r 2 . Sector area is the fraction of circle area for given angle.