Equality of two chords formed by intersections of circle with tangent lines to another circle

Question

We have two circle C(O,OC) and C(Q,QA). We know that AO and BO are tangent for C(Q,QA) and QC and QD are tangent for C(O,OC). We want to prove that LJ = KT.

Is it possible to help me?

Answer 1

Let's call $OQ=d$, $OC=r_1$, $AQ=r_2$, $\angle AOQ=\theta_1$,$\angle CQO=\theta_2$. We have $\sin \theta_1=r_2/d$ and $\sin \theta_2=r_1/d$. Now $LJ=2r_1\sin \theta_1$ and $KT=2r_2\sin \theta_2$, but $$r_2\sin \theta_2=r_2\cdot \frac{r_1}{d}=r_1\cdot \frac{r_2}{d}=r_1\sin \theta_1=\frac{r_1r_2}{d}$$ so we have $KT=LJ=\frac{2r_1r_2}{d}$

Answer 2

Let $d$ be the distance between the center of the circles and $R$ and $r$ their respective radii .

$$OL=OJ$$ (they are equal radii) and also $$OA=OB$$ (they are tangents) .

Now it's obvious that $\triangle OLJ $ and $\triangle OAB$ are similar so :

$$\frac{LJ}{AB}=\frac{OL}{OA}$$

$OL=R$

$OA=\sqrt{d^2-r^2}$

Also $$AB=2 \cdot \frac{OA \cdot AQ}{OQ}=\frac{2r\sqrt{d^2-r^2}}{d}$$

Plugging we find that $LJ=\frac{2Rr}{d}$ which is symmetrical in $R$ and $r$ so also $KT=\frac{2Rr}{d}=LJ$ as wanted .