Which of the following is a point of tangency on the circle below?

Mathematics

1 answer:

lesya [120]2 years ago

5 0

Answer:

Answer D but read below.

Step-by-step explanation:

The point of tangency is the endpoint of either a radius or a diameter such that only one point of the circumference of the circle is concurrent with the endpoint of the diameter or radius of the circle.

Technically, it is none of them. E should be 180 degrees. That is really a sloppy question.

However I think you are expected to answer F which is answer D.

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The slope-intercept form of the equation of a line that passes through point (-3, 8) is y = -2/3x + 6. What is the point-

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$y = \stackrel{\stackrel{m}{\downarrow }}{-\cfrac{2}{3}}x+6\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so hmmm then we know the slope of that line is -2/3, so we're really looking for the point-slope form of a line with a slope of -2/3 and that passes through (-3 , 8)

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{8})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{8}=\stackrel{m}{-\cfrac{2}{3}}(x-\stackrel{x_1}{(-3)})\implies y-8=-\cfrac{2}{3}(x+3)$