# Circle Properties and Missing Values

Please help describe how to solve problems based on circle properties and circle rules.

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Solve problems and justify arguments about chords and lines tangent to circles

Let's take a quick review on a few parts of a circle.

Circle

The locus of a point, which moves such that its distance from a fixed point always a constant.

The fixed point is called its center and the constant distance is called its radius.

The boundary of the circle is called its circumference.

Chord

A line segment whose end points lie on the circumference of the circle is called a chord.

AB is the chord.

Let's review the properties related to the chord.

Property 1:

In a circle, chords equidistant from the center are congruent.

Property 2:

In a circle, congruent chords are equidistant from the center.

Property 3:

A perpendicular line drawn from the center of a circle to a chord bisects the chord.

Property 4:

The line joining the center and the midpoint of a chord is perpendicular to the chord.

Tangent

A tangent to a circle is a line in the plane of the circle that intersects the circle in exactly one point.

The line m is called the tangent line.

Point of tangency

The point where a circle and a tangent intersect is the point of tangency.

The point A is called the point of tangency

.

Now let's review the properties related to the tangent lines.

Property1:

If a line is tangent to a circle, then the line is perpendicular to the radius to the point of tangency.

Property 2:

If a line in the plane of a circle is perpendicular to a radius at its endpoints on the circle, then the line is tangent to the circle.

AB is tangent to the circle O.

Property 3:

The two segments tangent to a circle from a point outside the circle are congruent.

That is, TP = TQ.

Let's work out problems related to the chords and tangents to the circles.

Example

Find the length of the missing side OA when AB = 8 cm and OC = 3 cm.

Given : AB = 8 cm and OC = 3 cm

To find: Length of OA.

Property: Perpendicular from the center of a circle to a chord bisects the chord.

==> AC = CB = 4 cm [ since AB = 8 cm]

Let's find the length of OA using the Pythagorean theorem.

In âˆ†OCA, OA2 = OC2 + CA2

OA2 = 32 + 42

= 9 + 16

= 25

OA = 5 cm

The length of the missing side OA is 5 cm.

Example

Find the value of x.

Given: OC = 18 and AC = 25

To find: Length of DE.

1)The perpendicular drawn from O to DE is F which equals 18.

==> OC = OF = 18 [Given:Perpendicular drawn from the center to the chords are equal]

2) AC = CB [Property:Perpendicular drawn from the center bisects the chord]

==> AB = AC + CB

AB = 25 + 25 = 50

3) AB = DE [Property: Chords equidistant from the center are congruent]

50 = DE

The length of the chord DE is 50.

Example

Find the length of OC.

Given: AC = CB = 18 and DE = 36

To find : Length of OC

The chords AB and DE have same length.

i.e AB = DE = 36

==> OF = OC [Property:Equal chords of a circle are equidistant from the center]

12 = OC

The length of OC is 12.

Example

Find the measure of the angle P.

Given: PQ and PR are the tangents.

To find: m<P

OQ and OR are the radius of the circle perpendicular to the tangents PQ and PR.

==> <Q and <R are the right angles.

PQOR is a quadrilateral whose angle measures have a sum of 360 degrees.

m<P + m<Q + m<R + m<O = 360Â°

m<P + 90Â° + 90Â° + 122Â° = 360Â°

m<P + 180Â° + 122Â° = 360Â° [90Â° + 90Â° = 180Â°]

m<P + 302Â° = 360Â° [180Â° + 122Â° = 302Â°]

m<P = 360Â° - 302Â°

m<P = 58Â°

The measure of the angle P is 58Â°.

Example

Determine the perimeter of the triangle ABC.

Given: AF = 12 cm , DB = 14 cm, and CE = 7 cm

To find: Perimeter of the triangle ABC.

The circle with center O is inscribed in the triangle ABC.

AD = AF = 12 cm

[Property: The two segments tangent to a circle from a point outside the circle are congruent]

BD = BE = 14 cm

CF = CE = 7 cm

We know that perimeter is the sum of the side lengths of the figure.

Perimeter of the triangle ABC = AD + DB + BE + EC + CF + FA

= 12 + 14 + 14 + 7 + 7 + 12

= 66

The perimeter of the triangle ABC is 66 cm.

AB2 = EB2 + AE2

AB2 = 52 + 262

AB2 = 25 + 676

AB2 = 701

AB = 26.47

The distance between the gears is about 26.5 in.

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