3 years ago

Pls answer! this problem

Mathematics

1 answer:

sleet_krkn [62]3 years ago

5 0

There isn’t a question your asking?

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Consider the circle of radius 5 centered at (0,0), how do you find an equation of the line tangent to the circle at the point (3

Flauer [41]

$\text{We know that the tangent at any point on a circle is perpendicular}\\
\text{to the radius at that point.}\\
\\
\text{so first we find the equation of the line joining (0,0) and (3,4)}\\
\text{and then using the slope, we find the line perperndicular to it at (3,4)}\\
\\
\text{the equation of the line passing through (0,0) and (3,4) is}\\
\\
y-0=\frac{4-0}{3-0}(x-0)$

$\Rightarrow y=\frac{4}{3}x\\
\\
\text{so the slope of radius is }m=\frac{4}{3}.\\
\\
\text{we know that the product of the perpendicular lines is }-1. \\
\\
\text{so the slope of the perpendicular line would be}=-\frac{1}{4/3}=-\frac{3}{4}\\
\\
\text{So the equation of the tangent line has slope }-\frac{3}{4} \text{ and}\\
\text{passing through (3,4). so equation of tangent line is}$

$y-4=-\frac{3}{4}(x-3)\\
\\
\Rightarrow y-4=-\frac{3}{4}x+\frac{9}{4}\\
\\
\Rightarrow y=-\frac{3}{4}x+\frac{9}{4}+4\\
\\
\Rightarrow y=-\frac{3}{4}x+\frac{9+16}{4}\\
\\
\text{so the equation of tangent line is:} y=-\frac{3}{4}x+\frac{25}{4}$