Why are confidence intervals of hazard ratios not symmetric? My understanding is that the confidence interval for a hazard ratio should be symmetrical about the mean (the distance between the lower limit and the mean is the same as the distance between the mean and the upper limit). But often when I see a hazard ratio in the published literature on clinical trials this is not the case.
For example, a study reporting HR, 0.69; 95% CI, 0.54 to 0.89 in mCRC for cetuximab plus FOLFOX-4 vs FOLFOX-4 alone found here: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7044820/pdf/bmjopen-2019-030738.pdf.
The distance between the lower limit and the mean is 0.69 - 0.54 = 0.15, while the distance between the mean and the upper limit is 0.89 - 0.69 = 0.20. Shouldnt the distance for both be 0.15 to be symmetric?

The log (partial) likelihood maximization methods used to fit survival models work in the log-hazard scale. The regression coefficient estimates are in that scale, and in that scale the covariance of the estimates is assumed to be multivariate normal. In the log-hazard scale, confidence intervals (CI) are thus symmetric.
When you exponentiate those coefficients and corresponding CI limits to get results in terms of hazard ratios (HR), the CI are necessarily asymmetric. That's even true for the usual null-hypothesis assumption of log-hazard-ratio of 0 or HR of 1. For example, if the log-hazard-ratio estimate has CI of (-0.2, +0.2) around a point estimate of 0, in the (exponentiated) HR scale the CI are (0.8187, 1.2214), asymmetric about the point estimate of HR = 1.
More broadly, as other answers nicely demonstrate, there is no need for CI to be symmetric at all. They often are taken to be symmetric when coefficient estimates can be assumed to have symmetric distributions, but even then there is no rule requiring such a choice. Any interval containing 95% of the probability distribution can be taken as 95% CI.
