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I'm totally lost with this question. I appreciate any kind of help.

if the equation of a circle is $(x-3)^2+y^2=9$

Find :

-Equation of the tangent line at $(2,2\sqrt2)$

-Equation of the tangent to the circle symmetric about the $x$ axis to the line obtained in the first question.

From the equation I found that :

Center is $(3,0)$ Radius $= 3$

Thanks for your help

Edit : I can't use calculus or per-calculus to solve it

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  • $\begingroup$ The reason people were using calculus to solve this problem is that you tagged it linear algebra, for which calculus is a prerequisite, rather than algebra-precalculus. Please change the tag. $\endgroup$ – N. F. Taussig Sep 28 '14 at 11:00
  • $\begingroup$ I understand , my bad $\endgroup$ – User2010101 Sep 28 '14 at 17:12
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Since you know the center and the point of tangency, you can compute the slope of the radius to the point of tangency. Since the center is $(3, 0)$ and the point of tangency is $(2, 2\sqrt{2})$, the slope of the radius to the point of tangency is

$$m_r = \frac{2\sqrt{2} - 0}{3 - 2} = 2\sqrt{2}$$

The slope of the tangent line to the circle is perpendicular to the radius at the point of tangency. If two non-vertical lines are perpendicular, their slopes are negative reciprocals, so the slope of the tangent line to the circle at $(2, 2\sqrt{2})$ is the negative reciprocal of the slope of the radius to the point of tangency. Thus, the tangent line has slope

$$m_{\perp} = -\frac{1}{2\sqrt{2}} = -\frac{1}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{4}$$

You can then use the point-slope equation

$$y - y_0 = m(x - x_0)$$

to write the equation of the tangent line, where $(x_0, y_0)$ is the point $(2, 2\sqrt{2})$ and $m$ is the slope of the tangent line. The equation of the tangent line to the circle $(x - 3)^2 + y^2 = 9$ at $(2, 2\sqrt{2})$ is

$$y - 2\sqrt{2} = -\frac{\sqrt{2}}{4}(x - 2)$$

A reflection in the $x$-axis sends point $(x, y)$ to the point $(x, -y)$. Thus, the reflection of the point $(2, 2\sqrt{2})$ in the $x$-axis is $(2, -2\sqrt{2})$. To find the equation of the tangent line to the circle at this point, follow the steps outlined above with $(2, -2\sqrt{2})$ replacing $(2, 2\sqrt{2})$.

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  • $\begingroup$ Can you do it step by step? I'm still lost. Thanks! $\endgroup$ – User2010101 Sep 28 '14 at 3:25
  • $\begingroup$ I demonstrated how to find the equation of the tangent line to the circle at $(2, 2\sqrt{2})$. If you follow the same procedure, you should now be able to find the equation of the tangent line at $(2, -2\sqrt{2})$. $\endgroup$ – N. F. Taussig Sep 28 '14 at 11:03
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since the tangent passes through the point $T=(2,2\sqrt{2})$ we need to find its slope. this is the value of $\frac{dy}{dx}$ at $T$.

differentiating the equation gives: $$ 2(x-3) + 2y\frac{dy}{dx} = 0 $$ at any point. setting $x=2$ and $y=2\sqrt{2}$ allows calculation of $m_T=\frac{dy}{dx}|_T$

the tangent is then: $$ y-2\sqrt{2} = m_T(x-2) $$

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  • $\begingroup$ Thanks for your reply , but I can't use calculus or per-calculus to solve it $\endgroup$ – User2010101 Sep 28 '14 at 2:32
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Hint: Note that any line tangent is perpendicular to radio, then $$m_1m_2=-1$$

If $m_1$ is the slope of the radio in $(2,2\sqrt{2})$, then $m_1=\frac{2\sqrt{2}-0}{2-3}=-2\sqrt{2}$. Now if $m_2$ is the slope of the tangent, then $$m_1m_2=-1\Longrightarrow-2\sqrt{2}m_2=-1\Longrightarrow m_2=\frac{\sqrt{2}}{4}$$ You find the equation of the tangent line: $$y-2\sqrt{2}=m_2(x-2)$$

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  • $\begingroup$ I would appreciate if you help me with the example , because is the only way I really learn. $\endgroup$ – User2010101 Sep 28 '14 at 2:36
  • $\begingroup$ Ok I edite just now. $\endgroup$ – AsdrubalBeltran Sep 28 '14 at 3:06
  • $\begingroup$ Can you do it step by step? I'm still lost. Thanks! $\endgroup$ – User2010101 Sep 28 '14 at 3:23

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