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My maths teacher told me this problem was impossible without knowledge of implicit differentiation: is she right?

You are given the equation of the circle $\left(x+2\right)^2+\left(y-2\right)^2=16$ , what are the equations of the lines with a gradient of 2 which are tangents of this circle.

Here is a visualisation of the problem enter image description here

The red circle is the one I described, and the green and blue lines are the equations I'm looking for.

I think the solution will have something to do with plugging in the value of 2 in some sort of equation for the gradient i.e. $2=\frac{x-x_0}{y-y_0}$ , and then the discriminant the equation you would use to find where they intersect can be solved to find values of $x$ , but I'm not sure how to find these values without using any implicit differentiation -can it be done?

EDIT:

Ok so now I'll try and run this through step by step until i find another problem:

So we can say that there's a line (the diameter) of gradient -1/2 which passes through the centre (-2,2):

$y=mx+c$

$y=-\frac{1}{2}x+c$

$2=1+c$

$y=-\frac{1}{2}x+1$

We can also say that a solution lies on the point $(a,b)$ such that

$\left(a+2\right)^2+\left(b-2\right)^2=16$ and $b=-\frac{1}{2}a+1$

So we can substitute b in the equation of the circle to arrive at

$\left(a+2\right)^2+\left(-\frac{1}{2}a-1\right)^2=16$

which simplifies down to

$\frac{5}{4}a^2+5a-11=0$

and solves to give $\frac{2}{5}\left(-5-4\sqrt{5}\right)$ and $\frac{2}{5}\left(4\sqrt{5}-5\right)$

so these are the possible values for $x$

Now if i substitute 1 of these back into the equation, I'll get two values of $y$ for the one value of $x$ because it's quadratic, but there's only 1 intersection point at each value of $x$ , what do i do at this point?

EDIT 2:

It looks like both the intersection points of the line $x=\frac{2}{5}\left(4\sqrt{5}-5\right)$ correspond to the $y$ values of the intersection points of the tangents, does this mean i only need to plug one of these $x$ values back into the equation? Visualisation to show you what i mean (the blue line appears to intersect the circle at both the appropriate $y$ values)

enter image description here

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  • $\begingroup$ Do you the know the answer . I got $y=2x+7,y=2x-11$ $\endgroup$ – Archis Welankar Jun 5 '16 at 11:00
  • $\begingroup$ I think you need to find the equation of the diameter that meets the two lines you are searching for at the tangent points. That line passes through the center of the circle (-2,2) and its slope is -0.5. You can continue from here... $\endgroup$ – Noam Dolovich Jun 5 '16 at 11:03
  • $\begingroup$ From trial and error the answer is $y=2x+6+4\sqrt{5}$ or $y=2x+6-4\sqrt{5}$ , working it out fully now. $\endgroup$ – Cubbs Jun 5 '16 at 11:31
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    $\begingroup$ @Cubbs "Trial and error"?? How? Do you check all the pointsw on the circle? I don't think this would be well accepted in most schools. $\endgroup$ – DonAntonio Jun 5 '16 at 11:42
  • $\begingroup$ Just estimated it to 3 d.p. using a graphing calculator, then wolfram alpha can give you possible closed forms, one of which was that surd :) $\endgroup$ – Cubbs Jun 5 '16 at 11:54
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Since a tangent to a circle is perpendicular to the circle's radius at that points, we're looking for points $\;(a,b)\;$ on the circle such that the radius through them has a gradient (slope) of $\;-\frac12\;$ , so: as the circle's center is $\;(-2,2)\;$ , the slope from this point to $\;(a,b)\;$ is

$$\frac{b-2}{a+2}=-\frac12\implies -2b+4=a+2$$

but we also have

$$(a+2)^2+(b-2)^2=16\implies16=(-2b+4)^2+(b-2)^2=5(b-2)^2\implies$$

$$b-2=\pm\frac4{\sqrt5}\implies b=2\pm\frac4{\sqrt5}$$

and there you have the $\;y\,-\,$ coordinates of the wanted points. Now find out the $\;x\,-\,$ coordinates and that's all, since once you have the points you can alreayd write down the lines through them as you already have the slope (gradient): $\;2\;$

No calculus at all needed...though, perhaps, it could make things slightly simpler and faster.

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$$ (x+2)^2 + (y-2)^2 = 16 $$ $$ \Rightarrow x^2 + y^2+4x-4y+4+4 = 16 $$ $$ \Rightarrow x^2 + y^2+4x-4y-8= 0 $$ $$ \text{centre} =(-2,2), \text{ radius}=4 $$ for tangent the line must satisfies the condition, $r=d$ we have line $y=2x+c$ here gradient$=2$ need to find $c=$? $$ \pm4 =\frac {2(-2)-2+c}{2^2+1} $$ $$ \pm 20 =-6 +c$$ $$ c= 26,-14$$ now the tangents are $$y=2x+26, y=2x-14$$ are required tangents

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