As shown in the diagram, each circle's radius is 1, AG is tangent to O3, what's the length of EF?

Question

I've been stuck on this problem for awhile now. I can't figure out how to use Power of a Point to get EF, and the only lengths I have are AG, GO3, AO3, and that $AE\cdot AF=8$. Please help me out, and thank you!

Answer 1

Let $M$ be the midpoint of chord $EF$. $O_2M \perp AG$ and $O_3G\perp AG$.

By similar triangles $\triangle O_2AM \sim \triangle O_3AG$,

$$\begin{align*} \frac{O_2 M}{O_3G} &= \frac{AO_2}{AO_3}\\ O_2M &= \frac 35 \end{align*}$$

Then $EF$ is

$$\begin{align*}EF &= 2 EM\\ &= 2\sqrt{O_2E^2 - O_2M^2}\\ &= 2\sqrt{1^2 - \left(\frac 35\right)^2}\\ &= \frac 85 \end{align*}$$

Answer 2

I will solve it using trigonometry: AG is tangent to o3 OG⫠AG Angle o3ag= sin^(-1)⁡〖1/5〗 We’ll draw o2E and from the law of cosines: 1=9+(AE)^2 -6AE cos⁡sin^(-1)⁡〖1/5〗 (AE)^2 -6AE cos⁡sin^(-1)⁡〖1/5+8=0〗 So AE(6 cos⁡〖sin^(-1)⁡〖1/5〗-√((-6 cos⁡sin^(-1)⁡〖1/5〗 )^2-4×8)〗)/2=2.13 Note That 3.73 is rejected. So 2.13(2.13+EF)=8 So EF = 1.6 (I used the ANS)