Advanced Circles - Circles - High School Geometry Unlocked (2016)

High School Geometry Unlocked (2016)

Chapter 7. Circles

GOALS

By the end of this chapter, you will be able to:


•Understand tangent lines, secant lines, and chords, and the relationships among these and other parts of a circle

•Calculate the length of an arc, or the area of a sector of a circle

•Solve problems using the ratios of arc length : circumference, sector area : total area, and central angle : total angle

•Understand the relationships between inscribed and central angles

•Solve problems with concentric circles

•Understand radians, and convert between radians and degrees

Lesson 7.1. Advanced Circles

SECANTS AND CHORDS

If we have a triangle inscribed in a circle as shown above, with two vertices on the circle’s circumference and one vertex at its center, then the triangle is certainly isosceles. We know this because all radii in a circle are equal. In the figure, two of the triangle’s sides are radii of the circle; therefore, they are congruent.

In Lesson 5.1,
we reviewed
radius, diameter,
circumference, and
area. In this chapter,
you’ll build on your
knowledge of circles,
and the relationships
among different
parts of a circle.

A circle has infinitely many triangles that can be formed in this manner.

The line segment that connects two points on a circle is called a chord. If that segment is extended to become a line, that line is called a secant. If a chord passes through the center of a circle, then it is a diameter.

If a radius is perpendicular to a chord, then it bisects the chord. The converse is also true—if a radius bisects a chord, then it is perpendicular to the chord.

Find the length of AB.

In this figure, the two halves of AB are congruent, since they are bisected by a perpendicular radius. If they are congruent, that means that we can set them equal to each other:

3x + 4 = 5x − 6

3x + 10 = 5x

Add 6 to both sides of the equation.

 10 = 2x

Subtract 2x from both sides of the equation.

 5 = x

We’ve concluded that x = 5. But don’t forget to solve for AB! Plug in 5 for x in both expressions, and add them together.

3x + 4 + 5x − 6

= 3(5) + 4 + 5(5) − 6

= 15 + 4 + 25 − 6

= 19 + 25 − 6

= 44 − 6

= 38

The length of AB is 38.

TANGENTS

A line tangent to a circle intersects the circle at exactly one point. The intersection point is known as the point of tangency. The radius intersects the point of tangency at a 90° angle.

Consider a triangle that is tangent to a circle, as shown in the right-hand figure above. One vertex is at the center of the circle, and the other two vertices are on the tangent line. The triangle is not necessarily isosceles; however, we do know a little bit more than that. If we connect a radius at the point of tangency, the radius would be perpendicular to the tangent line, and in fact, the radius would serve as an altitude of the triangle.

Here is how you may see chords on the ACT.

In the circle below, radius OP is 10 inches long, ∠LOP is 60°, and OP is perpendicular to chord LN at M. How many inches long is LN?

A.10

B.3

C.10

D.5

E.5

A circle has infinitely many triangles that can be formed in this manner.

In the figure above, line l is tangent to the circle with center O at point P. What is the value of x?

If line l is tangent to the circle, then we know that line l is perpendicular to the radius shown. That means that there is only one unknown angle, x, in the triangle. To solve for x, subtract the two known angles from 180°.

x = 180° − 90° − 25°

x = 90° − 25°

x = 65°

If two tangent lines intersect, then each has the same length from the intersection point to the point of tangency.

Start by labeling the length of OP and the measurement of ∠LOP on the figure. LO is another radius of the circle, so label that with a length of 10 inches as well. △LMO is a 30-60-90 triangle, with ∠MLO equal to 30°. LO is opposite the 90° angle, soLO is the 2x side. If 2x = 10, then x = 5. MO is opposite the 30° angle, so MO is the x side and is equal to 5. LM is opposite the 60° angle, so it is the x side and is equal to 5. This is part of the length of LN, but the length of MN still needs to be found. Draw in the radius connecting points N and O to create another triangle. Because LO and NO are the same length and LN is perdendicular to OP, the new triangle is congruent to△LMO. Therefore, MN is equal to LM, and LN equals 5 + 5or 10. Choice (A) is the credited response.

In the figure above, lines l, m, and n are each tangent to the circle with center O. What is the perimeter of the triangle formed by lines l, m, and n?

From the rule above, we know that two intersecting tangent lines will form congruent segments. Find the segments that are congruent in this figure.

For the purposes of this exercise, we’ve labeled the vertices and tangent points in the figure. Segments AB and AF are intersecting tangent segments, so we know that they are congruent. Therefore, AF has a length of 5.

For the same reasons, we know that BC is congruent to CD, and that DE is congruent to EF. Therefore, CD has a length of 3 and EF has a length of 2.

Add up all the lengths to find the perimeter:

2 + 2 + 3 + 3 + 5 + 5 = 20

The triangle’s perimeter is 20.