Circle diameter endpoints to radius (SAT). Game on.

Circle diameter endpoints to radius (SAT) is a grade 10 math skill aligned to Common Core standard HSG.GPE.A.1: derive the equation of a circle of given center and radius using the Pythagorean Theorem; complete the square to find the center and radius of a circle given by an equation. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 circle diameter endpoints to radius (sat) problems our math games drill.

CCSS HSG.GPE.A.110 questions in the bank
Kickoff

8 plays. No signup.

Warm-upeasy

A circle in the coordinate plane has a diameter with endpoints (−6,−5)(-6, -5) and (−6,9)(-6, 9). An equation of the circle is (x+6)2+(y−2)2=r2(x + 6)^2 + (y - 2)^2 = r^2, where r>0r > 0. What is the value of rr?

The diameter is vertical with length ∣9−(−5)∣=14|9 - (-5)| = 14, so r=7r = 7.

Mid-gameeasy

The endpoints of a diameter of a circle are (2,−4)(2, -4) and (18,−4)(18, -4). The circle has equation (x−10)2+(y+4)2=r2(x - 10)^2 + (y + 4)^2 = r^2. Find the positive constant rr.

The diameter is horizontal with length ∣18−2∣=16|18 - 2| = 16, so r=8r = 8.

Mid-gameeasy

A circle in the coordinate plane has a diameter with endpoints (8,0)(8, 0) and (8,6)(8, 6). An equation is (x−8)2+(y−3)2=r2(x - 8)^2 + (y - 3)^2 = r^2. What is the value of rr?

The diameter is vertical with length ∣6−0∣=6|6 - 0| = 6, so r=3r = 3.

Mid-gameeasy

The endpoints of a diameter of a circle are (−6,1)(-6, 1) and (12,1)(12, 1). The circle has equation (x−3)2+(y−1)2=r2(x - 3)^2 + (y - 1)^2 = r^2. Find rr.

The diameter is horizontal with length ∣12−(−6)∣=18|12 - (-6)| = 18, so r=9r = 9.

Mid-gameeasy

A circle in the coordinate plane has a diameter with endpoints (2.5,−1)(2.5, -1) and (2.5,7)(2.5, 7). An equation of the circle is (x−2.5)2+(y−3)2=r2(x - 2.5)^2 + (y - 3)^2 = r^2. What is the value of rr?

The diameter is vertical with length ∣7−(−1)∣=8|7 - (-1)| = 8, so r=4r = 4.

Mid-gamemedium

Given diameter endpoints (0,0)(0, 0) and (6,8)(6, 8) and equation (x−3)2+(y−4)2=r2(x - 3)^2 + (y - 4)^2 = r^2, what is rr?

The diameter length is (6−0)2+(8−0)2=36+64=100=10\sqrt{(6 - 0)^2 + (8 - 0)^2} = \sqrt{36 + 64} = \sqrt{100} = 10. Then r=102=5r = \dfrac{10}{2} = 5.

Mid-gamemedium

Given diameter endpoints (1,−3)(1, -3) and (13,13)(13, 13) and equation (x−7)2+(y−5)2=r2(x - 7)^2 + (y - 5)^2 = r^2 in the coordinate plane, what is rr?

The diameter length is (13−1)2+(13−(−3))2=122+162=144+256=400=20\sqrt{(13 - 1)^2 + (13 - (-3))^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20. Then r=10r = 10.

Buzzer beatermedium

Given diameter endpoints (−5,2)(-5, 2) and (5,26)(5, 26) and equation x2+(y−14)2=r2x^2 + (y - 14)^2 = r^2, what is the positive constant rr?

The diameter length is (5−(−5))2+(26−2)2=102+242=100+576=676=26\sqrt{(5 - (-5))^2 + (26 - 2)^2} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26. Then r=13r = 13.

Tests that cover circle diameter endpoints to radius (sat):
NAEPSATACTNY State MathNC EOG/EOC
More Grade 10 skills
Advanced area formulasCircle equation translationsNormal distribution basicsUnit circle trig values (SAT)Circle theoremsEquation of a circleMidpoint formulaPerimeter and arc lengthAll Grade 10 practice →

Drill it inside a game.

The free placement test finds your level, then every match serves circle diameter endpoints to radius (sat) questions at exactly the right difficulty.