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Being $f(x)=\sqrt{1-(x-2)^2}$ I have to know what linear equation only touches the circle once(only one intersection), and passes by $P(0,0)$.

So the linear equation must be $y=mx$ because $n=0$.

I have a system of 2 equations: \begin{align} y&=\sqrt{1-(x-2)^2}\\ y&=mx \end{align}

So I equal both equations and I get \begin{align} mx&=\sqrt{1-(x-2)^2}\\ m&=\frac{\sqrt{1-(x-2)^2}}{x} \end{align}

$m$ can be put in the $y=mx$ equation, which equals to: \begin{align} y&=\left(\frac{\sqrt{1-(x-2)^2}}{x}\right)x\\ &=\sqrt{1-(x-2)^2} \end{align}

But that equation has $\infty$ intersections, and I want only the equation who has $1$ interception.

What is the good way to know this? And how can it be calculated?

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As you have in your post, we have $y = mx$ as the straight line. For this line to touch the semi-circle, we need that $y = mx$ and $y = \sqrt{1 - (x-2)^2}$ must have only one solution. This means that the equation $$mx = \sqrt{1 - (x-2)^2}$$ must have only one solution.

Hence, we need to find $m$ such that $m^2x^2 = 1 - (x-2)^2$ has only one solution.

$(m^2 + 1)x^2 -4x +4 = 1 \implies (m^2+1) x^2 - 4x + 3 = 0$.

Any quadratic equation always has two solution. The two solutions collapse to a single solution when the discriminant of the quadratic equation is $0$. This is seen from the following reasoning.

For instance, if we have $ax^2 + bx+c = 0$, then we get that $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$ i.e. $\displaystyle x_1 = \frac{-b + \sqrt{b^2 -4ac}}{2a}$ and $\displaystyle x_2 = \frac{-b - \sqrt{b^2 -4ac}}{2a}$ are the two solutions. If the two solutions to collapse into a single solution i.e. if $x_1 = x_2$, we get that $$ \frac{-b + \sqrt{b^2 -4ac}}{2a} = \frac{-b - \sqrt{b^2 -4ac}}{2a}$$ This gives us that $\displaystyle \sqrt{b^2 -4ac} = 0$. $D = b^2 - 4ac$ is called the discriminant of the quadratic.

The discriminant of the quadratic equation, $(m^2+1) x^2 - 4x + 3 = 0$ is $D = (-4)^2 - 4 \times 3 \times (m^2+1)$.

Setting the discriminant to zero gives us that $(-4)^2 - 4 \times 3 \times (m^2 + 1) =0$ which gives us $ \displaystyle m^2 + 1 = \frac43 \implies m = \pm \frac1{\sqrt{3}}$.

Hence, the two lines from origin that touch the circle are $y = \pm \dfrac{x}{\sqrt{3}}$.

Since you have a semi-circle, the only line that touches the circle is $\displaystyle y = \frac{x}{\sqrt{3}}$. (Thanks to @Joe Johnson 126 for pointing this out).

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  • $\begingroup$ What did you do to get $(-4)^2 - 4 \times 3 \times (m^2 + 1) =0$ from $(m^2 + 1)x^2 -4x +4 = 1 \implies (m^2+1) x^2 - 4x + 3 = 0$? $\endgroup$ – Garmen1778 May 21 '12 at 20:35
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    $\begingroup$ @Marvis: The OP does not have a circle. So, $y=-\frac{1}{\sqrt{3}}x$ does not touch the graph of $f(x)=\sqrt{1-(x-2)^2}$. $\endgroup$ – Joe Johnson 126 May 21 '12 at 20:37
  • $\begingroup$ @JoeJohnson126 Thanks for pointing this out. I have updated the post accordingly. $\endgroup$ – user17762 May 21 '12 at 20:44
  • $\begingroup$ @Garmen1778 I have added more details to the post. $\endgroup$ – user17762 May 21 '12 at 20:44
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You were doing fine up through $mx=\sqrt{1-(x-2)^2}$. Starting there, square both sides to get $m^2x^2=1-(x-2)^2$, and rewrite this as $(m^2+1)x^2-4x+3=0$. Think of this as a quadratic in $x$; solving yields

$$x=\frac{4\pm\sqrt{16-12(m^2+1)}}{2(m^2+1)}=\frac{2\pm\sqrt{4-3(m^2+1)}}{m^2+1}\;,$$

which has two solutions or none unless $4-3(m^2+1)=0$, i.e., $m^2=\frac13$, and $m=\pm\frac1{\sqrt3}$.

However, the equation $f(x)=\sqrt{1-(x-2)^2}$ describes only the upper half of the circle with centre $\langle 2,0\rangle$ and radius $1$, so the line through the origin must have positive slope in order to intersect the semicircle. Its equation is therefore $$y=\frac{x}{\sqrt3}\;.$$

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Draw a picture. Our curve is the top half of the circle with centre $(2,0)$ and radius $1$.

Draw the tangent line to the curve that passes through the origin. Suppose it meets the half-circle at $P$. Let $C=(2,0)$ be the centre of the half-circle. Then $CP$ is perpendicular to $OP$, and therefore the sine of $\angle COP$ is $CP/OP$, which is $1/2$. So the tangent of $\angle COP$ is $\dfrac{1}{\sqrt{3}}$, and therefore the tangent line has equation $y=\dfrac{x}{\sqrt{3}}$.

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