# Finding the equations of all lines tangent to the circle $x^2+y^2=2y$ and passing through $(0, 4)$

I can't figure out this question:

Find the equations of all lines that are tangent to the circle $$x^2 + y^2 = 2y$$ and pass through the point $$(0, 4)$$.

Hint: The line $$y = mx + 4$$ is tangent to the circle if it intersects the circle at only one point.

Things I've tried: I've tried things from making a right angled triangle where $$4$$ is the hypotenuse, $$\sqrt{2y}$$ being $$a^2$$ or $$b^2$$ and try to solve for distance that way, then after trying to get the distance from $$(0,4)$$ to the unknown point of the tangent on the circle which I will call $$(x,y)$$ which yielded no results

I've also tried to equate the gradients $$m_1 m_2 = -1$$ but after graphing this circle out I believe the center was not $$(0,0)$$ as the equation $$x^2 + y^2 = 2y$$ implied (even if it was $$(0,0)$$ I still can't figure it out).

My graph of how the question might work

• Have you tried to use the hint? Plugging $y = mx+4$ into $x^2+y^2=2y$ should give you a quadratic equation. Now ask yourself, when does a quadratic equation have exactly one solution? Oct 11, 2020 at 13:52
• Thanks Klaus I managed to solve it i think. You the MVP Oct 11, 2020 at 15:42

The line $$mx-y+4=0$$ is tangent to the circle if the distance from its center is equal to the radius.

The center is $$(0,1)$$ and the radius is $$r=1$$.

so we must have $$\frac{|-1+4|}{\sqrt{m^2+1}}=1$$ square both sides $$\frac{9}{m^2+1}=1$$ $$m=\pm 2\sqrt{2}$$ Tangent equations are $$y=2x\sqrt2+4 ;\;y=-2x\sqrt 2 +4$$

Differentiating $$x^2+y^2=2y$$, we get$$2x+2y\frac{dy}{dx}=2\frac{dy}{dx}\implies x=\frac{dy}{dx}(1-y)$$ $$\frac{dy}{dx}=\frac{x}{1-y}$$

Ok with Klaus's help I managed to figure it out so I will do it for the future ppl who googling this question a favor. many thanks to Klaus!

$$x^2+y^2=2y$$

$$y=mx+4$$

$$x^2+m^2x^2+8mx+16 = 2mx + 8$$

$$x^2+m^2x^2+6mx+8 = 0$$

$$(1+m^2)x^2+6mx+8$$, with a = ($$1+m^2$$), b = $$6m$$, c = $$8$$

$$\sqrt{b^2-4ac} = 0$$

$${(6m)^2-4(1+m^2)(8)} = 0$$

$$m = \sqrt{8}$$