Circles

SAT Questions that focus on Circles require knowledge of the following topics.

Equation of a Circle in the Cartesian Coordinate System

In the Cartesian Coordinate System, the equation of a circle 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.

Length of the Circumference

1 radian is defined in a circle as the angle whose arc is equal to one circle's radius (following figure).

Since one full circle has $2\pi$ radians, the length of the circumference with radius "1" is $2\pi$. And, since the length is proportional to the radius, the length of a circumference with radius "R" is: $2{\pi}R$.

π (pi) is an irrational constant approximately equal to 3.14159.

Area of a Circle

The area of a circle with radius "R" is given by the formula: $A_C={\pi}R^2$

Lines Tangent to a Circle

Consider the following figure, where $t_1$ and $t_2$ are two tangents to the circle with center $O$ and radius $r$:

The two triangles $AT_1O$ and $AT_2O$ are congruent, because they share the same hypotenuse (AO) and both have one cathetus that measures $r$ (the radius of the circle). Therefore:
- angles $T_1\hat{A}O$ and $T_2\hat{A}O$ are congruent, and 
- segments $AT_1$ and $AT_2$ are the same length.

Circumscribed Quadrilaterals

Consider the quadrilateral ABCD circumscribed to the circle in the following figure:

In a circumscribed quadrilateral, the sum of the lengths of two opposite sides is equal to the sum of the lengths of the two other opposite sides: $AB+DC=AD+BC$.

As seen previously (Lines Tangent to a Circle), we know that:

$A1=A4$
$B1=B2$
$C3=C2$
$D3=D4$

Adding the four equalities:

$(A1+B1)+(C3+D3)=(A4+D4)+(B2+C2)$
$AB+DC=AD+BC$

Inscribed Angles

In the following figure, triangle ABC is inscribed in the circle with radius "r" and center "O":

Triangle AOC is isosceles, since it has two sides of equal length, "r". Thus the two base angles of triangle AOC are congruent, "x".
Triangle AOB is also isosceles, since it has two sides of equal length, "r". Thus the two base angles of triangle AOB are also congruent, "y".

The sum of the three internal angles of triangle ABC is equal to 180º. Thus:
$y+y+x+x+O\hat{B}C+O\hat{C}B=180$
(I) $O\hat{B}C+O\hat{C}B=180-2x-2y$

The sum of the three internal angles of triangle BOC is equal to 180º, Thus:
(II) $B\hat{O}C+O\hat{B}C+O\hat{C}B=180$

Plugging I into II:
$B\hat{O}C+180-2x-2y=180$
$B\hat{O}C=2x+2y=2(B\hat{A}C)$

Angle $B\hat{O}C$ is called central angle;
Angle $B\hat{A}C$ is called inscribed angle.

Conclusion: The inscribed angle is half the central angle that subtends the same arc on the circle (it doesn't matter where on the circle the vertex is placed).

Inscribed Quadrilaterals

Quadrilateral ABCD is inscribed in the following circle:

Angles $A\hat{B}C$ and $C\hat{D}A$ are inscribed angles. Their corresponding central angles subtend arcs ABC and ADC. The sum of the measures of these two central angles is 360º, that is, a full circle. Thus,

$A\hat{B}C+C\hat{D}A=180º$.

Conclusion: opposite angles in inscribed quadrilaterals are supplementary.

Interior Angle (or Internal Eccentric Angle)

Consider the following figure:

The two chords AC and BD intersect in point P, forming angle $r$, called interior angle.

Angle $A\hat{B}D$ is an inscribed angle. Thus, it measures half of arc AD ($A\hat{B}D=\frac{\overset{\frown}{AD}}{2}$).

Angle $B\hat{A}C$ is also an inscribed angle. Thus, it measures half of arc BC ($B\hat{A}C=\frac{\overset{\frown}{BC}}{2}$).

Since angle $r$ is an external angle of triangle ABP, it measures the same as the sum of the two opposite internal angles of this triangle, that is,

$r=A\hat{B}D+B\hat{A}C$

$r=\frac{\overset{\frown}{AD}}{2}+\frac{\overset{\frown}{BC}}{2}$.

Note that an inscribed angle is one particular case of the interior angle, a case in which one of the arcs is equal to 0º.

Exterior Angle

Consider the following figure:

The two segments AP and DP intersect in point P (outside the circle), forming an angle $r$, called exterior angle.

Angle $A\hat{B}D$ is an inscribed angle. Thus, it measures half of arc AD ($A\hat{B}D=\frac{\overset{\frown}{AD}}{2}$).

Angle $B\hat{D}C$ is also an inscribed angle. Thus, it measures half of arc BC ($B\hat{D}C=\frac{\overset{\frown}{BC}}{2}$).

Since angle $A\hat{B}D$ is an external angle of triangle DBP, it measures the same as the sum of the two opposite internal angles of this triangle, that is,

$A\hat{B}D=r+B\hat{D}C$

$\frac{\overset{\frown}{AD}}{2}=r+\frac{\overset{\frown}{BC}}{2}$

$r=\frac{\overset{\frown}{AD}}{2}-\frac{\overset{\frown}{BC}}{2}$

Segment Relationships in Circles

In the following figure, segments AC and BD are secant segments that intersect inside the circle, $r$ is an interior angle, and $s$ and $t$ are inscribed angles:

Triangles APD and BPC are congruent (congruent internal angles), thus

$\frac{PD}{PC}=\frac{PA}{PB}$

$(PD)(PB)=(PA)(PC)$, that is, when two secant segments intersect inside a circle, the product of the segment pieces of one segment is equal to the product of the segment pieces of the other.


In the following figure, segments AB and CD are secant segments that intersect outside the circle, $r$ is an exterior angle, and $s$ is an inscribed angle:

Triangles APC and DPB are congruent (congruent internal angles), thus

$\frac{PA}{PD}=\frac{PC}{PB}$

$(PA)(PB)=(PD)(PC)$

If one of the segments is tangent to the circle (PD for example), $PD=PC$, and the right side of the equation would be $(PD)^2$ (only when one of the segments is tangent to the circle!).

Solved SAT Practice Tests

Find Practice Tests in the following link:

SAT Practice Tests - Circles and

Additional Practice Tests - Circles