# Parabola and Circle problem : The parabola $y =x^2-8x+15$ cuts the x axis at P and Q. A circle is drawn …

Problem :

The parabola $y=x^2-8x+15$ cuts the x axis at P and Q. A circle is drawn through P and Q so that the origin is outside it. Find the length at a tangent to the circle from O.

My approach :

Since the parabola $y=x^2-8x+15$ cuts the x axis therefore, its y coordinate is zero,

Solving the equation: $x^2-8x+15=0$ we get two points $(3,0)$ and $(5,0)$.

Now how to proceed further with these two points, please suggest. thanks..

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There are infinite circles that pass through $\,P,Q\,$ and do not contain the origin within them. Is this what you really meant? Because if so the wanted length is something that can't be given numerically but only parametrically, perhaps as a function of the circle's radius and/or center... – DonAntonio Aug 29 '13 at 10:59
@DonAntonio See my answer. – Mark Bennet Aug 29 '13 at 11:13
Yes @MarkBennet, what I said: a parametric answer. – DonAntonio Aug 29 '13 at 11:22
@DonAntonio Look carefully - the parameter cancels – Mark Bennet Aug 29 '13 at 11:23
Indeed so. Nice. +1 – DonAntonio Aug 29 '13 at 11:29

Where is the centre of the circle - at some point

$C=(4,a)$

What is the square of the radius of the circle:

$r^2=a^2+1$

What is the square of the distance from the origin to the centre of the circle:

$OC^2=4^2+a^2$

Let $S$ be a point on the circle where the tangent from the origin touches it. We have a right-angled triangle with $OS^2+CS^2=OC^2$ and we know that $CS^2=r^2$

Can you finish it from there?

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Note that the expression for $OC^2$ implies you can't have a circle through those two points which has the origin inside it. – Mark Bennet Aug 29 '13 at 11:12

Let $T$ be the point of tangency. By the Power of a Point "tangent-secant" theorem

\begin{align} |\overline{OT}|^2 &= |\overline{OP}|\;|\overline{OQ}| \\ &= 3 \cdot 5 \end{align} so that $$|\overline{OT}| = \sqrt{15}$$

Others have observed that, although there are infinitely-many circles through $P$ and $Q$, the length of the tangent segment from $O$ is always the same. Consequently, the set of all those points of tangency forms a circle around $O$, and this circle is said to be "orthogonal" to each member of the infinite family: it crosses each member at right angles. Such configurations have importance in more-advanced geometry.

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Perhaps I'm missing something here but this is the second time in the last 1-2 weeks Iread this power of a point thingy in this site, and this seems to be only the renaming of several well-known basics theorem in euclidean geometry, like the products of parts of two intersecting cords/secants to a a circle (which follow painfully easy from similarity considerations), or the length of a tangent to a circle (Pythagoras), etc. The name also seems to be used mainly in english speaking countries. – DonAntonio Aug 29 '13 at 12:03
@DonAntonio: I refer to the Power of a Point as often as I can; it's one of my favorite results. :) Arguably, it's "only the renaming of several well-known basic theorems", but the reason for the renaming is that those theorems assign a numerical value (the so-called "power") to any point relative to a given circle. (The power value is the squared-length of the tangent for external points, or the negative product of parts of intersecting chords for internal points.) (continued) – Blue Aug 29 '13 at 12:34
(Part 2) The conceptual significance to me is that the theorem just about comparing parts of these two chords (or secants or tangents) that happen to intersect at some point $P$. (As you mention, similarity easily covers that.) Rather, the significance is that every chord/secant/tangent through $P$ exhibits a certain numerical relationship based on an intrinsic numerical property of the point (relative to circle). That's a (ahem) power-ful concept. (BTW: I should mention that the power itself is $d^2-r^2$, where $r$ is the circle's radius, and $d$ the distance from point to center.) – Blue Aug 29 '13 at 12:44
(Part 3) So, for instance, in the context of the OP's question, we have that the origin has the same power with respect to every circle in the infinite family through $P$ and $Q$. This is a very concise way of imparting info about how any line through $O$ interacts w/those circles. Even so ... More often than not, I tend to invoke the Power of a Point theorem mundanely, for ad hoc comparisons of pairs of chords/secants/tangents. It's almost insulting to what the Power of a Point means, but if casual mention of the theorem gets people curious about it, then perhaps I've done a service. – Blue Aug 29 '13 at 13:05
pretty nice and explicative. Thanks.+1 – DonAntonio Aug 29 '13 at 13:23

In the brown right-angled triangle, we can see that:$$r^2=y_c^2+1$$ In the right-angled triangle formed on the x-axis with OR as its hypotenuse, we see that:$$m^2=y_c^2+4^2=y_c^2+16$$ Finally, in the right-angled triangle ORT, we see that:$$l^2=m^2-r^2=(y_c^2+16)-(y_c^2+1)$$ Hopefully you can finish off from here.

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Your picture seems to be upside-down... Are you familiar with GeoGebra? – Servaes Aug 29 '13 at 11:23
Oops - sorry - my mistake - thanks for pointing it out @DonAntonio. :) – Mufasa Aug 29 '13 at 11:23
I have corrected the picture - thanks again – Mufasa Aug 29 '13 at 11:28

Note that we can use the general theorem that if $OT$ is a tangent to a circle and $OPQ$ is a straight line meeting the circle at points $P$ and $Q$ we have $OT^2=OP\cdot OQ$

[this can be proved using similar triangles given that the angle between $OT$ and $TP$ is equal to the angle between $OQ$ and $TQ$ (given $OQ\gt OP$)]

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That's an aspect of the Power of a Point Theorem, as mentioned in my answer. :) – Blue Aug 29 '13 at 11:34