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3 years ago

prove that if two circles are tangent externally then the common internal tangent bisects a common external tangent.

Mathematics

1 answer:

Scilla [17]3 years ago

5 0

Explanation:

Consider the attached diagram.

Circles A and B have mutual tangents CF and DE that intersect at point F.

We know that the two tangents to a circle from an external point are congruent, so ...

FE≅ FC

FD ≅ FC

By the transitive property of congruence, both segments congruent to FC are congruent to each other:

FD ≅ FE

Therefore, CF is a bisector of DE.

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Answer:

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Step-by-step explanation:

We are given the following expression and we are to solve it for the variable x:

$x ^2 + \frac { 1 } { 2 } x + \frac { 1 } { 1 6 } = \frac { 4 } { 9 }$

We will find the least common multiple of 2. 6 and 9:

$x^2 \times 144 +\frac{1}{2}x \times 144 +\frac{1}{16} \times 144 = \frac{4}{9} \times 144$

Simplifying it to get:

$144x^2+72x+9=64$

$144x^2+72x+9-64=0$

$144x^2+72x-55=0$

Using the quadratic formula to solve for x:

$x=\frac{-b \pm\sqrt{b^2-4ac} }{2a}$

$x=\frac{-72 \pm\sqrt{72^2-4(144)(-55)} }{2a}$

$x=\frac{5}{12}$ or $0.417$

$x=-\frac{11}{12}$ or $-0.917$

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