circlestangent line
Its obvious (by intuition) that from an external point(outside a circle) only two tangents can be drawn to the circle.
But , how do we prove it prove it?
N.B: 1. Please provide the simplest possible proof.
2. Here , it is proved for a parabola. But , how can we do the same for a circle.
First note that the quadrilaterial $PAOB$ is a kite, this follows from the fact that the tangent from one point to a circle are from a same length, and the $AO=OB=r$.
From the fact that $\angle AOB = 60^{\circ}$ and using $AO=OB$ we get that the $\triangle AOB$ is equilaterial. This leads to conclusion $AB=r=1$, which is the smaller diagonal from the kite. Now we need to find the other one.
Let $D$ be the point when both diagonals intersect, and from the propertiy of the kite we know that they are normal to each other. So we can split the diagonal $PO$ into $PO = OD + PD$. We know that the $OD$ is the height of the equilaterial triangle so we have:
$$OD = \frac{r\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$
We know that the angle between the tangent line to a circle and the radius at the touching point is $90^{\circ}$, so $\angle PAO = 90^{\circ}$. We can denote $\angle PAO$ as sum of two angles: $\angle PAO = \angle OAB + \angle BAP = 60^{\circ} + \angle BAP$, because $\angle OAB$ is an angle in an equilaterial triangle. So from this we have:
$$\angle BAP = 30^{\circ}$$
Also we know that $AD = \frac r2 = \frac 12$. Now in the right triangle $PAD$ we have:
$$\tan \angle BAP = \frac{PD}{AD}$$ $$\tan \angle 30^{\circ} = \frac{PD}{\frac 12}$$ $$ \frac {1}{\sqrt{3}} = \frac{PD}{\frac 12}$$
$$ PD = \frac{1}{2\sqrt{3}}$$
Now we have: $PO = OD + PD$, i.e.:
$$PO = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{3}{2\sqrt{3}} + \frac{1}{2\sqrt{3}}$$ $$PO = \frac{4}{2\sqrt{3}}$$ $$PO = \frac{2}{\sqrt{3}} \approx 1.154$$
So the distance from $O$ to $P$ is constant, that means that $P$ can be any point that's $\frac{2}{\sqrt{3}}$ units away from $(0,0)$, i.e. $P$ lies on a circle with this equation:
$$x^2 + y^2 =\left(\frac{2}{\sqrt{3}}\right)^2$$ $$x^2 + y^2 = \frac 43$$
So the locus of the point $P$ is:
$$x^2 + y^2 = \frac 43$$
There can be infinitely many tangents to a curve (not always ofcourse). For eg. consider $y=\sin x$ and the line $y=1$. It touches the curve at infinitely many points. For the fig given assume the external pt. is $(4,1)$. Then, the $y=1$ is a tangent to infinite number of points on the $\sin x$ curve through that point.
Yes, a tangent at a point can cut the curve somewhere else. Let,$y=x\sin x$. Each tangent for a local maxima cuts the curve somewhere else.
Interesting read: A tangent at a point of inflection can cut right through the curve rather than just touching it. Do some research on it, if interested.
For eg.
Best Answer
Without the loss of generality say the circle is $x^2+y^2=r^2$. Let $P(a,b)$ be the external point. A line passing through $P$ with slope $m$ is of the form $$y-b=m(x-a).$$
To get the intersection points of this line and the circle, we substitute $y$ in the equation of the circle and obtain $$x^2(m^2+1)+2mx(b-am)+((am-b)^2-r^2)=0.$$
For tangency, we want equal roots. Thus we want the discriminant to be $0$. This gives us $$4m^2(b-am)^2-4(m^2+1)((am-b)^2-r^2)=0.$$ Upon simplifying we get, $$m^2r^2-(am-b)^2-r^2=0.$$ This is a quadratic equation in $m$ so at the most two real values. This means there can be only two possible tangent lines emanating from $P$.
In fact, it can be easily shown that if this equation has a real root then it will have two distinct real roots, thus exactly two tangents.