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 (3,4)?

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Answer 1 · 2015-04-20T19:19:19

Using calculus:

the equation of the circle is $x^{2} + y^{2} = 25$ $x^2+y^2 = 25$

At the point $(3, 4)$ $(3,4)$ , we are on the upper semicircle, whose equation is: $y = \sqrt{25 - x^{2}}$ $y = sqrt(25-x^2)$

So $\frac{d y}{d x} = \frac{1}{2 \sqrt{25 - x^{2}}} (- 2 x) = \frac{- x}{\sqrt{25 - x^{2}}}$ $dy/dx = 1/(2sqrt(25-x^2)) (-2x) = (-x)/sqrt(25-x^2)$

At $(3, 4)$ $(3,4)$ , we get $\frac{d y}{d x} = \frac{- 3}{4}$ $dy/dx = (-3)/4$

So the slope of the line is $- \frac{3}{4}$ $-3/4$ and the line contains the point $(3, 4)$ $(3,4)$ .

$y = - \frac{3}{4} x + \frac{25}{4}$ $y=-3/4 x +25/4$

Without calculus:

The tangent is perpendicular to the radius connecting $(0, 0)$ $(0,0)$ and $(3, 4)$ $(3,4)$ .

The slope of the radius is $\frac{4}{3}$ $4/3$ , so the slope of the tangent is $- \frac{3}{4}$ $-3/4$