Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 11.1
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Exercise 44 Page 687

For a given point and a circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle.

Area: 16Ď€ square units
Circumference: 8 π units

Practice makes perfect

We want to find the area and the circumference of the circle C from the following diagram.

In order to do so, we should first find its radius or diameter. Let's find its radius.

Radius

For a given point and a circle, the product of the lengths of the two segments from the point to the circle is constant along any line through the point and circle.

In our diagram, the point which will follow this rule is the point of intersection of the shown chord segments. Therefore, the products of the lengths of the chord segments are equal. x * 2=4 * 3 Let's solve this equation for x.
x * 2 = 4 *3
2x=12
x = 6
Now, notice that the green segment passes through the center C of the circle. This means that its length equals the diameter. Radius is half this length. r = x+2/2 ⇒ r = 6+2/2 = 4

Area

Now that we know the radius of the circle, we can calculate its area using the following formula. A = π r^2 Let's substitute 4 for r to find the area.
A = π r^2
A = π ( 4)^2
A = π(16)
A = 16Ď€
The area is 16Ď€ square units.

Circumference

We also need to find the circumference of the given circle. Let's recall the formula for it. C = 2 π r Again, let's substitute 4 for r and simplify.
C = 2 π r
C = 2 π ( 4)
C = 8 π
The circumference is 8Ď€ units.