What is the slope of the tangent to the graph of x^2 + y^2 = 9 that intersects the x-axis at (4, 0).

Expert Answers

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Let the line that is tangential to the graph x^2 + y^2 = 9 intersect it at the point (x1, y1).

The slope of the line joining (x1, y1) and (4, 0) is (y1 - 0)/(x1 - 4)

The slope of the line can also be derived by using the derivative of x^2 + y^2 = 9.

2x + 2y*y' = 0

=> y' = -x/y

At the point (x1, y1), the slope of the tangent is -x1/y1

Equating the slope derived using the two methods we get:

(y1 - 0)/(x1 - 4) = -x1/y1

=> y1^2 = -x1^2 + 4x1

Also y1^2 + x1^2 = 9

=> 4x1 = 9

=> x1 = 9/4

y1^2 = 9 - (9/4)^2

=> y1 = -(sqrt 63)/4 and (sqrt 63)/4

The slope of the tangent can be 9/sqrt 63 or -9/sqrt 63

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