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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

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    $\begingroup$ 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? $\endgroup$
    – Klaus
    Oct 11, 2020 at 13:52
  • $\begingroup$ Thanks Klaus I managed to solve it i think. You the MVP $\endgroup$ Oct 11, 2020 at 15:42

3 Answers 3

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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 $$y=2x\sqrt2+4 ;\;y=-2x\sqrt 2 +4$$

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

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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 = $8$

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

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

$m = \sqrt{8}$

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