What are the two values of $k$ such that the line $y=kx-4$ crosses $x^2+y^2=4$ at one distinct point?

Question

I am quite lost here, I am pretty much choosing the slope of the line that hits the circle sitting above it. Any advice on how to proceed? I thought to replace $y$ in the equation of the circle with $kx-4$ giving me $$x=\frac{4k\pm 2\sqrt{k^2-3}}{1+k^2}$$ but I am not sure how or if this is helpful?

Any advice would be amazing…

Answer 1

There's a very important part of the question that you're missing: one distinct point. When you have a quadratic such as $$x^2+(kx-4)^2=4$$

it's common to find two distinct solutions for $x$ (as you did), but here there's only one repeated root. And in a quadratic equation, repeated roots occur when the discriminant $b^2-4ac$ that you see in the quadratic formula is equal to zero. So if we expand out the above quadratic we get

$$x^2+k^2x^2-8kx+16=4$$

which, in standard form, is

$$(1+k^2)x^2-(8k)x+12=0$$

which has discriminant

$$b^2-4ac=64k^2-4(1+k^2)(12)=16k^2-48$$

$$\implies 16k^2-48=0$$

so $16k^2=48$, or $k= \pm \sqrt{3}$.

Answer 2

Here's another approach to take using the geometrical properties of a circle. We seek the two lines through $ \ (0 \ , \ -4) \ $ that are tangent to the circle $ \ x^2 + y^2 \ = \ 4 \ \ , $ which is centered on the origin and has a radius of $ \ 2 \ \ . $ At the points where the tangent lines contact the circle, the radius to that point is perpendicular to the tangent line. If we draw in these radii, we have two right triangles with hypotenuses of length $ \ 4 \ \ $ and the short "leg" of each has length $ \ 2 \ \ . $ We can then use the "Pythagorean Theorem" to find the length of the long "legs " extending from $ \ (0 \ , \ -4) \ $ to the tangent points.

The slope of the tangent lines can be found from the ratio of the length of the vertical violet line segment to the length of the horizontal violet line segment; those segments form the "legs" of right triangles with the lengths we found in the first paragraph being the hypotenuses. These new right triangles are similar to the right triangles from the first paragraph, so we have $$ \ k \ = \ \frac{y}{x} \ = \ \frac{\text{long leg}}{2} \ \ ; $$ taking into account the sign of $ \ x \ $ for each of the two violet triangles, we will have two values for $ \ k \ \ . $