Question

prove algebraicially that the straight line with equation x=2y+5 is tangent to a circle with equation x²+y²=5

Answer 1

The given line is tangent to the circle at the point (-2, 1).

How to prove that the line is tangent to the circle?

Here we have the circle equation:

$x^2 + y^2 = 5$

Writing that as a function (actually two functions) we get:

$x = \pm\sqrt{5 - y^2}$

If we differentiate the part with positive sign, we get:

$\frac{dx}{dy} = (-1/2)*(5 - y^2)^{-1/2}*(-2y)$

Now, in which value of y we need to evalute that to get a 2?

If we use y = -2 we get:

$\frac{dx}{dy} = (1/2)*(5 - 2^2)^{-1/2}*(-2*-2) = 2$

now, when y = -2 in the circle we get:

$x^2 + (-2)^2 = 5\\x^2 = 5 - 4 = 1\\x = \pm 1$

This means that the tangent line of the form:

x = 2y + b

Needs to pass through the point (-2, 1) or (-2, -1) to be tangent to the circle.

When evaluating in -2 we get:

x = 2*(-2) + b

x = -4 + b

Then we can have either b = 3 or b = 5.

So the line:

x = 2x + 5

Is tangent to the circle at the point (-2, 1)

If you want to learn more about tangent lines:

https://brainly.com/question/9636512

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