I am trying to solve 2 questions

Is there a way to find an equation for the length of a chord between two tangents in terms of the radius and distance of that chord to the external point?

Is there a way to find an equation for the distance between that chord and the center of the circle in the same terms (radius and distance of chord to external point)?

Please show working/proof

Here is a graphical representation

Lets define some terms:

$w = AB$

$t = OI$

$h = IP$ ie height of $ΔABP$

$r = OB$ ie radius

$AP$ and $BP$ are tangents to the circle

I am wanting to find:

  • An equation for $w$ in terms of $h$ and $r$ ie "$w = ...$"

  • An equation for $t$ in terms of $h$ and $r$ ie "$t = ...$"

Is this actually possible??

I have had a go and concluded the following:

$ΔBOP|||ΔAOP$ (radius, shared side + 90deg) (congruent)

$ΔBOP☰ΔIBP$ (shared angle, 3 angles) (similar)

$ΔIBP☰ΔIOB$ ($∠IOB=∠IBP$ [Angle of chord to tangent is half angle of sector created by chord], 3 angles) (similar)


But I don't no where to go from there


UPDATE So using the similar triangles $\Delta IOB \sim \Delta IBP$ we can state: $$ \frac{w}{t}=\frac{h}{w}, \therefore w^2 = th $$

We can also use Pythagoras Theorem on $\Delta IOB$ $$ \frac{w}{t}=\frac{h}{w}, \therefore w^2 = th $$

  • 1
    $\begingroup$ Could you format your mathematical symbols? $\endgroup$ – user122049 Dec 22 '17 at 7:51
  • $\begingroup$ I did some, wasn't sure how to do it specifically but is that ok? $\endgroup$ – SamofWise Dec 22 '17 at 12:26
  • $\begingroup$ You might have mentioned that you’re asking about tangents to a circle. Obvious from the diagram, of course, but seeing that requires chasing a link. $\endgroup$ – amd Dec 22 '17 at 21:08

The hint.

Since $BL$ is an altitude of $\Delta OBP$ and $\measuredangle OBP=90^{\circ},$ we obtain $$w^2=th$$ and $$w^2=h\sqrt{r^2-w^2}.$$ Now, solve the last equation.

  • $\begingroup$ So im not sure how you get the second equation. Ps thanks for fast reply. $\endgroup$ – SamofWise Dec 22 '17 at 12:32
  • $\begingroup$ If $\measuredangle ACB=90^{\circ}$ and $CD$ is an altitude of $\Delta ABC$ then $CD^2=AD\cdot DB$ and use the Pythagoras theorem: $t=\sqrt{r^2-w^2}.$ $\endgroup$ – Michael Rozenberg Dec 22 '17 at 12:34
  • $\begingroup$ Awesome so that leaves us with something like $w^4-h^2w^2-h^2r^2=0$. So does that mean the only way to solve it is using the quadratic formula? Ie $let x=w^2$ Solve: $x^2-h^2x-h^2r^2=0$ $\endgroup$ – SamofWise Dec 22 '17 at 12:45
  • $\begingroup$ Yes, of course. And we'll get an unique value of $w$. $\endgroup$ – Michael Rozenberg Dec 22 '17 at 12:47
  • $\begingroup$ Cool, wish it was cleaner but it works $\endgroup$ – SamofWise Dec 22 '17 at 12:48

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