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Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent to the circle at the point (3, 4) in slope intercept form.

User Ueli
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1 Answer

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Answer:


\displaystyle y= -(3)/(4) x + (25)/(4).

Explanation:

The equation of a circle of radius
5 centered at
(0,0) is:


x^(2) + y^(2) = 5^(2).


x^(2) + y^(2) = 25.

Differentiate implicitly with respect to
x to find the slope of tangents to this circle.


\displaystyle (d)/(dx)[x^(2) + y^(2)] = (d)/(dx)[25]


\displaystyle (d)/(dx)(x^(2)) + (d)/(dx)(y^(2)) = 0.

Apply the power rule and the chain rule. Treat
y as a function of
x,
f(x).


\displaystyle (d)/(dx)(x^(2)) + (d)/(dx)(f(x))^(2) = 0.


\displaystyle (d)/(dx)(2x) + (d)/(dx)(2f(x)\cdot f^(\prime)(x)) = 0.

That is:


\displaystyle (d)/(dx)(2x) + (d)/(dx)\left(2y \cdot (dy)/(dx)\right) = 0.

Solve this equation for
\displaystyle (dy)/(dx):


\displaystyle (dy)/(dx) = -(x)/(y).

The slope of the tangent to this circle at point
(3, 4) will thus equal


\displaystyle (dy)/(dx) = -(3)/(4).

Apply the slope-point of a line in a cartesian plane:


y - y_0 = m(x - x_0), where


  • m is the gradient of this line, and

  • (x_0, y_0) are the coordinates of a point on that line.

For the tangent line in this question:


  • \displaystyle m = -(3)/(4),

  • (x_0, y_0) = (3, 4).

The equation of this tangent line will thus be:


\displaystyle y - 4 = -(3)/(4) (x - 3).

That simplifies to


\displaystyle y= -(3)/(4) x + (25)/(4).

Consider the circle of radius 5 centered at (0, 0). Find an equation of the line tangent-example-1
User Huppo
by
8.0k points

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