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?


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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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


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 $$y=\frac{x}{\sqrt3}\;.$$


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 $y=\dfrac{x}{\sqrt{3}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.