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

MATH110 December 2010

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[hide] 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).

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[show] 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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[show] Solution
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: $y+4={\frac {3}{4}}(x-3)$