Secants and Tangents

[neno89]

Hello I really need some help. I'm not sure of how to do the following problems. I need to know how to do these problems for a mid term test on thursday. Can some one help me??!!

1.

B is 34 mm from the center of circle O, which has a radius 16 mm. Segment BP and segment BR are tangent segments. Line AC is tangent to circle O at point Q. Find the perimeter of triangle ABC.


2.

Given: Two concentric circles with center E; segment AB is tangent at C; Segment CD perpendicular to segment AE; AB = 40
Find: AF


3.

Segment BC is tangent to circle A at B, and segment BD is congruent to segment BA, Explain why segment BD bisects segment AC.


4.

Circles P and Q are tangent to each other and to the axes as shown. PQ = 26 and AB = 24. Find coordinates of P and Q.

 

[soroban]

Hello, neno89!

Here's #1 . . . I'll try to explain it without a new diagram.

1. B is 34 mm from the center of circle O, which has a radius 16 mm.
Segment BP and segment BR are tangent segments.
Line AC is tangent to circle O at point Q.
Find the perimeter of triangle ABC.

Draw radii \(\displaystyle OP\) and \(\displaystyle OR\); their length is 16.
Draw \(\displaystyle BO\,=\,34\).

\(\displaystyle BP\) is tangent to the circle at \(\displaystyle P.\;\)Hence, \(\displaystyle OP\,\perp\,BP\).
\(\displaystyle BR\) is tangent to the circle at \(\displaystyle R.\;\)Hence, \(\displaystyle OR\,\perp\,BR\).
\(\displaystyle \;\;\)Hence, \(\displaystyle \Delta BPO\) and \(\displaystyle \Delta BRO\) are right triangles.

Using Pythagorus: \(\displaystyle \,BP^2\,+\,OP^2\:=\:BO^2\;\;\Rightarrow\;\;BP^2\,+\,16^2\:=\:34^2\;\;\Rightarrow\;\;BP^2\,=\,900\;\;\Rightarrow\;\;BP\,=\,30\)

Then: \(\displaystyle BP\,=\,BR\,=\,30\;\;\) (1)
\(\displaystyle \;\;\)(Tangents to a circle from an external point have equal lengths.)
And note that: \(\displaystyle AQ\,=\,AP\) and \(\displaystyle QC\,=\,RC\;\) (for the same reason). \(\displaystyle \;\) (2)


The perimeter of \(\displaystyle \Delta ABC\) is: \(\displaystyle \,P\;=\;BA\,+\,AQ\,+\,QC\,+\,CB\)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \downarrow\) . . . . \(\displaystyle \downarrow\)
. . . . . . . . From (2), we have: \(\displaystyle \\;=\:\underbrace{BA\,+\,AP}\,+\,\underbrace{RC\,+\,CB}\)

. . . . . . . . . . . . . . .But this is: \(\displaystyle \\;=\;\;\;\;\;BP\;\;\;+\;\;\;BR\)


And from (1), we have: \(\displaystyle \,P\;=\;30\,+\,30\:=\:60\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Fascinating!
We draw \(\displaystyle AC\) tangent to the circle at \(\displaystyle Q\).
No matter how we position that tangent,
\(\displaystyle \;\;\)the perimeter of \(\displaystyle \Delta ABC\) is always 60 mm.