Science:Math Exam Resources/Courses/MATH110/December 2010/Question 06 (b)

MATH110 December 2010

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Question 06 (b)
A circle of radius 5 centered on the origin has the equation $x^{2}+y^{2}=25$ Find the equation of the line tangent to the circle at the point (3, -4).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
It may help first to find an expression for the bottom half of the circle by solving the equation of the circle for y.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Our differentiation rules at this point only address functions that are in the form $y=f(x)$. Thus our first step is to transform the equation of a circle into this form by solving for y. ${\begin{aligned}x^{2}+y^{2}&=25\\y^{2}&=25-x^{2}\\y&=\pm {\sqrt {25-x^{2}}}\end{aligned}}$ Because our point of concern is on the lower half of the circle, we will use the negative square root. We now proceed to find the slope of the tangent line at (3, -4). First we take the derivative using the chain rule. $\displaystyle {\begin{aligned}f(x)&=-{\sqrt {25-x^{2}}}\\f'(x)&=-{\frac {1}{2}}(25-x^{2})^{-{\frac {1}{2}}}(-2x)\\&={\frac {x}{\sqrt {25-x^{2}}}}\end{aligned}}$ To find the slope, we plug x = 3 into the derivative to get: $\displaystyle f'(3)={\frac {3}{\sqrt {25-3^{2}}}}={\frac {3}{4}}$ We now have a slope, $3/4$, and a point, $(3,-4)$. Plugging these into our equation of a line we get: $y+4={\frac {3}{4}}(x-3)$

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