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Deffense [45]
3 years ago
6

Prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x^2+y^2=5

Mathematics
2 answers:
Kay [80]3 years ago
8 0
<h3>Explanation:</h3>

A tangent line will have a couple of characteristics:

  • there is exactly one point of intersection with the circle
  • a perpendicular line through the point of tangency intersects the center of the circle

Substituting for x in the equation of the circle, we have ...

  (2y+5)^2 +y^2 = 5

  5y^2 +20y +20 = 0 . . . . simplify, subtract 5

  5(y +2)^2 = 0 . . . . . . . . . factor

This equation has exactly one solution, at y = -2. The corresponding value of x is ...

  x = 2(-2) +5 = 1

So, the line intersects the circle in exactly one point: (1, -2).

__

The center of the circle is (0, 0), so the line through the center and point of intersection is ...

  y = -2x . . . . . . . . . slope is -2

The tangent line is ...

  y = 1/2x -5/2 . . . . . . slope is 1/2

The product of slopes of these lines is (-2)(1/2) = -1, indicating the lines are perpendicular.

__

We have shown ...

  • the tangent line intersects the circle in one point: (1, -2)
  • the tangent line is perpendicular to the radius at the point of tangency.

zmey [24]3 years ago
7 0

Differentiate both sides of the equation of the circle with respect to x, treating y=y(x) as a function of x:

x^2+y^2=5\implies2x+2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac xy

This gives the slope of any line tangent to the circle at the point (x,y).

Rewriting the given line in slope-intercept form tells us its slope is

x=2y+5\implies y=\dfrac12x-\dfrac52\implies\mathrm{slope}=\dfrac12

In order for this line to be tangent to the circle, it must intersect the circle at the point (x,y) such that

-\dfrac xy=\dfrac12\implies y=-2x

In the equation of the circle, we have

x^2+(-2x)^2=5x^2=5\implies x=\pm1\implies y=\mp2

If x=-1, then -1=2y+5\implies y=-3\neq2, so we omit this case.

If x=1, then 1=2y+5\implies y=-2, as expected. Therefore x=2y+5 is a tangent line to the circle x^2+y^2=5 at the point (1, -2).

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