Prove algebraically that the straight line with equation x=2y+5 is a tangent to the circle with equation x^2+y^2=5i have gotten 4/5 marks but i need to get 5/5 you!

Differentiate both sides of the equation of the circle with respect to , treating as a function of :

This gives the slope of any line tangent to the circle at the point .

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

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

In the equation of the circle, we have

If , then , so we omit this case.

If , then , as expected. Therefore is a tangent line to the circle at the point (1, -2).

A tangent line will have a couple of characteristics:

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).

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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.

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We have shown ...

i think 5 miles per hour

step-by-step explanation: