I may not be a lexicographer, but I do know what affixes are. An affix is a word element attached to either the beginning or ending of the root of a word with the purpose of modifying the original word's meaning. Here are a few examples of an affix.
Look at how many words, for example, build off of the root -clos-. This root, at its core, means to "shut." Here are a few I could think of:
Closet |
Closely |
Closest |
Disclose |
Foreclosure |
Closeness |
Closemouthed |
Notice how combinations of letters added to the word change the meaning of the word while still preserving the original meaning. Let's try an easier root, -spec-, which means "to see." This one has so many examples that I will not even attempt to list them all. I tried listing ones with unique affixes and lengths:
Retrospectively |
Specifications |
Disrespectful |
Spectrograph |
Spectacular |
Inspectors |
Suspected |
Spectrum |
Aspects |
Speck |
All of these words (and many others I omitted) are some variation of "to see."
I do not think this extreme diversity of words would be possible without affixes. Imagine if there was a completely different word for every variation of -spec- or -clos-? These are only two examples. You can attempt to investigate -aqua- or -jur- or -chrono-.
Despite this analysis, I am not sure whether or not it would change the number of words in a language. Maybe there is some other way of organizing symbols to create words that we are not implementing in the modern day; you have to start from somewhere. Maybe other people can chime in and provide their thoughts. I suspect that this is a toss-up question and that no definite answer exists.
A line is tangent to a circle if it intersects that circle at exactly one point. That's all we need to prove.
\(\boxed{1}\hspace{1mm}x=2y+5\\ \boxed{2}\hspace{1mm}x^2+y^2=5\) | We can use the method of substitution to find the point of intersection. |
\((2y+5)^2+y^2=5\) | We have to expand the square of a binomial in order to make some progress here. |
\(4y^2+20y+25+y^2=5\) | Combine the like terms and bring all terms to one side of the equation. |
\(5y^2+20y+20=0\) | Factor out the greatest common factor from every term, 5. |
\(5(y^2+4y+4)=0\) | The trinomial contained in the parentheses can be written as a square of a binomial. |
\(5(y+2)^2=0\) | Divide by 5 from both sides of the equation. |
\((y+2)^2=0\) | Take the square root of both sides. |
\(\sqrt{(y+2)^2}=\sqrt{0}\) | Write the left-hand side as the absolute value of the argument. |
\(|y+2|=0\) | Solve for y. |
\(y+2=0\) | |
\(y=-2\) | Now, solve for x. Since we are proving that this line is tangent, we should have foreseen that there would only be one possible y-value. Now, we must solve for the x-coordinate. |
\(\boxed{1}\hspace{1mm}x=2y+5; y=-2\) | Of course, you may substitute into either equation, but it is probably desirable to substitute the known y-value into this equation because it is most likely easier. |
\(x=-2*2+5\) | Simplify. |
\(x=1\) | |
Therefore, the intersection point of the line and circle is located at only \((1,-2)\). Since there is only one intersection point, the line is tangent to the circle.