Question

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

Answer 1

Explanation:

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.

Answer 2

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).