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.