Circles Pattern - Geometric Properties

This pattern tests your understanding of geometric relationships within circles — radii, chords, tangent lines, and inscribed shapes.

Chord Length from Triangle Perimeter

When two radii and a chord form a triangle, you can find the chord length from the perimeter.

A circular fountain has center $F$. Points $A$ and $B$ are on the edge. The radius is 12 feet. The perimeter of triangle $FAB$ is 40 feet. What is the chord length $AB$?

A) $12$
B) $16$
C) $12\sqrt{2}$
D) $16\sqrt{2}$

$FA$ and $FB$ are both radii, so each equals 12. Perimeter $= FA + FB + AB = 12 + 12 + AB = 40$. Therefore $AB = 16$. The answer is B.

This works because two sides of the triangle are radii (equal lengths), and the third side is the chord.

Two-Tangent Theorem

When two tangent lines are drawn from an external point $P$ to a circle with center $O$, the two tangent segments are equal in length: $PA = PB$.

Also, each tangent is perpendicular to the radius at the point of tangency: $OA \perp PA$ and $OB \perp PB$.

A circle with center $O$ has radius 140 cm. From external point $P$, tangent segments $PA$ and $PB$ touch the circle at $A$ and $B$. The perimeter of quadrilateral $OAPB$ is 1,240 cm. What is $OP$?

A) $140$
B) $480$
C) $500$
D) $620$

$OA = OB = 140$ (radii). By the two-tangent theorem, $PA = PB$. Perimeter $= OA + PA + PB + OB = 140 + PA + PA + 140 = 280 + 2 \cdot PA = 1{,}240$. So $PA = 480$.

Since $OA \perp PA$, triangle $OAP$ is a right triangle. By the Pythagorean theorem: $OP = \sqrt{OA^2 + PA^2} = \sqrt{140^2 + 480^2} = \sqrt{19{,}600 + 230{,}400} = \sqrt{250{,}000} = 500$. The answer is C.

A circle with center $O$ has radius 55 cm. Tangent segments from point $P$ create quadrilateral $OAPB$ with perimeter 530 cm. What is $OP$?

A) $55$
B) $210$
C) $217$
D) $265$

$PA = PB = \dfrac{530 - 2(55)}{2} = \dfrac{420}{2} = 210$. Then $OP = \sqrt{55^2 + 210^2} = \sqrt{3{,}025 + 44{,}100} = \sqrt{47{,}125} = 5\sqrt{1{,}885} \approx 217$. The answer is C.

Sector Area and Total Area

The minor sector and major sector together make up the full circle.

Points $X$, $Y$, and $Z$ are on a circle with center $O$. The minor sector $XOY$ has area 40 m$^2$ and the major sector (containing $Z$) has area 120 m$^2$. What is the total area of the circle?

A) $40$
B) $80$
C) $120$
D) $160$

Total $= 40 + 120 = 160$. The answer is D. The two sectors partition the entire circle.

Inscribed Isosceles Triangle Angles

When two sides of a triangle are radii of the same circle, the triangle is isosceles. The base angles are equal.

In a circle with center $O$, points $A$ and $B$ lie on the circle. The central angle $\angle AOB = 32°$. What is the measure of $\angle OAB$?

Triangle $OAB$ is isosceles with $OA = OB$ (radii). The base angles are equal: $\angle OAB = \angle OBA$. Since the angles of a triangle sum to $180°$: $32 + 2 \cdot \angle OAB = 180$, so $\angle OAB = 74°$.

In a circle with center $O$, the central angle $\angle AOB = 50°$. What is $\angle OAB$?

$\angle OAB = \dfrac{180 - 50}{2} = 65°$.

Key facts to remember:

Tangent segments from the same external point are equal in length. A tangent line is perpendicular to the radius at the point of tangency. Two radii to points on a circle form an isosceles triangle with the chord.

What to Do on Test Day

• Key facts to remember:
• Two radii to points on the circle form an isosceles triangle with the chord. The base angles are equal.
• Tangent segments from the same external point are equal in length.
• A tangent line is perpendicular to the radius at the point of tangency.
• Minor sector + major sector = full circle area.
• Chord from triangle perimeter: If two sides of a triangle inscribed in a circle are radii, they are equal. Subtract both from the perimeter to find the chord.
• Two-tangent problems: The perimeter of the kite-shaped quadrilateral $OAPB$ is $2r + 2 \cdot PA$. Solve for $PA$, then use the Pythagorean theorem on right triangle $OAP$ to find $OP$.
• Isosceles triangle angles: Central angle $+ 2 \times$ base angle $= 180°$. So base angle $= \dfrac{180° - \text{central angle}}{2}$.