SAT Practice Test - Math - Circles

SAT practice tests arranged by topic and difficulty level. In this section find tips and tactics for solving SAT questions that focus on Circles.

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SAT 2015 Practice Test.

In the figure above, point O is the center of the circle, line segments LM and MN are tangent to the circle at points L and N, respectively, and the segments intersect at point M as shown. If the circumference of the circle is 96, what is the length of minor arc LN?

Answer:

Segments LM and MN are tangent to the circle. Thus angles OLM and ONM are right angles.

The sum of the internal angles in a convex quadrilateral is 360º. Therefore:
$LON+90+60+90=360$
$LON+240=360$
$LON=120$ degrees.

The length of minor arc LN is proportional to angle LON. If the circumference of the whole circle (360º) is 96, than arc LN (120º) measures:

$\frac{360}{120}=\frac{96}{LN}$

$360LN=96(120)$
$3LN=96$
$LN=32$

Answer: 32

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SAT 2015 Test.

$x^2+y^2+4x-2y=-1$

The equation of a circle in the xy-plane is shown above. What is the radius of the circle?
A) 2
B) 3
C) 4
D) 9

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius.

Back to the given equation:
$x^2+y^2+4x-2y=-1$
$x^2+4x+y^2-2y=-1$
$x^2+4x+4-4+y^2-2y+1-1=-1$
$(x+2)^2-4+(y-1)^2-1=-1$
$(x+2)^2+(y-1)^2=-1+4+1$
$(x+2)^2+(y-1)^2=4$
$(x+2)^2+(y-1)^2=2^2$

Answer: A

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SAT 2015 Test. Which of the following is an equation of a circle in the xy-plane with center (0, 4) and a radius with endpoint (4/3, 5)?
A) $x^2+(y-4)^2=25/9$
B) $x^2+(y+4)^2=25/9$
C) $x^2+(y-4)^2=5/3$
D) $x^2+(y+4)^2=3/5$

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius.
Since the center of the given circle is (0, 4):
$(x-a)^2+(y-b)^2=R^2$
$(x-0)^2+(y-4)^2=R^2$
$x^2+(y-4)^2=R^2$

The radius is the distance between (0, 4) and (4/3, 5):
$R^2=(5-4)^2+(4/3-0)^2$
$R^2=(1)^2+(4/3)^2$
$R^2=1+16/9$
$R^2=25/9$

Therefore the equation of the circle is:
$x^2 + (y – 4)^2= R^2$
$x^2 + (y – 4)^2 = 25/9$

Answer: A

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SAT Practice Test.

The figure presents a metal nut with two hexagonal faces and six sides. The thickness of one side, from one hexagonal face to the other hexagonal face of the nut, is labeled as 1 centimeter.

The preceding figure shows a metal hex nut with two regular hexagonal faces and a thickness of 1 centimeter. The length of each side of a hexagonal face is 2 centimeters. A hole with a diameter of 2 centimeters is drilled through the nut. The density of the metal is 7.9 grams per cubic centimeter. What is the mass of this nut, to the nearest gram? (Density is mass divided by volume.)

Answer:

The following image is an hexagon with side 2cm:

An hexagon is composed by 6 equilateral triangles. In the figure above h is the hight of one of these tiangles. Let's calculate h using the Pythagoras Theorem in the rectangle triangle AMO:

$1^2+h^2=2^2$
$1+h^2=4$
$h^2=3$
$h=\sqrt{3}$

So the area of the equilateral triangle is $(1+1).h/2=\sqrt{3}$

The area of the hexagon is 6 times the area of each equilateral triangle: $A_{hexa}=6\sqrt{3}$.

The volume of the nut (before subtracing the volume of the hole) is the area of the hexagon times the thickness of the nut:
$V_{hexa}=6\sqrt{3}.1=6\sqrt{3}$

The volume of the hole is the area of the circle times the thickness of the nut:
$V_{hole}=\pi.1^2.1=\pi=3.14$

Therefore, the volume of the nut is:
$V_{hexa}-V_{hole}=6\sqrt{3}-3.14$
$V_{nut}=10.4-3.14=7,25cm^3$

The mass is the volume times the density:
$M_{nut}=7.25.(7.9)$
$M_{nut}=57,2$

Answer: 57g

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SAT Practice Test. $x^2+y^2-6x+8y=144$.

The equation of a circle in the x y plane is shown above. What is the diameter of the circle?

Answer:

The equation of a circle in the x y plane is given by: $(x – a)^2 + (y – b)^2 = R^2$, where (a, b) are the coordinates of the center of the circle, and R is its radius. Let's develop this equation:

$(x – a)^2 + (y – b)^2 = R^2$
$x^2-2ax+a^2+y^2-2by+b^2 = R^2$
$x^2-2ax+y^2-2by = R^2-a^2-b^2$
$x^2+y^2-2ax-2by = R^2-a^2-b^2$

Comparing this result to the given equation:
$-2ax=-6x$, so $a=3$
$-2by=8y$, so $b=-4$

And
$R^2-a^2-b^2=144$
$R^2-3^2-(-4)^2=144$
$R^2-9-16=144$
$R^2=144+9+16$
$R^2=169$
$R=13$

And the diameter (twice the radius) is 26.

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SAT Practice Test.

The figure presents a semicircle with a horizontal diameter labeled A B and a horizontal line segment labeled C D intersecting the semicircle at points C and D.

The semicircle above has a radius of r inches, and chord CD is parallel to the diameter AB. If the length of CD is 2/3 of the length of AB, what is the distance between the chord and the diameter in terms of r?

A. $\frac{1}{3}\pi.r$.

B. $\frac{2}{3}\pi.r$.

C. $\frac{\sqrt2}{2}.r$.

D. $\frac{\sqrt5}{3}.r$.

Answer:

In the following figure, O is the center of the circle, and M is the midpoint between C and D. The triangle OMD is rectangle:

Using the Pythagoras Theorem in triangle OMD:

$x^2+(\frac{2}{3}.r)^2=r^2$

$x^2+\frac{4}{9}.r^2=r^2$

$x^2=\frac{5}{9}.r^2$

$x=\frac{\sqrt5}{3}.r$

Answer: D

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