Another way to find the limits of the habitable zone of a star is to find the ratio between the star's luminosity and that of the Sun. Some sources give the luminosity of a star in units of solar luminosity. So divided the star's luminosity by that of the Sun the find the ratio.
The ratio will usually be either more than 1.00 or less than 1.00. In a few cases for stars very similar to the Sun, it will be 1.00 and maybe with many more zeroes.
Then find the square root of the ratio.
Then multiply the distance from the Sun of the inner edge of the Sun's circumstellar habitable zone by the square root of the ratio to find the distance from the star to the inner edge of the star's circumstellar habitable zone.
And do the same with the distance from the Sun to the outer edge of the Sun's circumstellar zone to find the outer edge of the star's circumstellar habitable zone.
The only flaw with that is that nobody really knows for 100 percent certain the inner edge of the Sun's circumstellar habitable.
Here is a link to a table of about a dozen scientific estimates or calculations of one or both edges of the Sun's circumstellar habitable zones.
https://en.wikipedia.org/wiki/Circumstellar_habitable_zone#Solar_System_estimates
They differ, sometimes by a lot.
The flaw in the formula used in James K's answer is that it is based on two of the calculations for the Sun's habitable zone, that of Kastings et al in 1993, and that Whitmire et al in 1996.
https://www.planetarybiology.com/calculating_habitable_zone.html
And you can see that some estimates of the Sun's habitable zone are much wider or narrower than Kasting's estimate.
It also links to a habitable zone calculator based on Kopparapu et al 2013.
https://live-vpl-test.pantheonsite.io/
Added just before midnight, March 11, 2023:
A "typical" G9V type star should have luminosity of 0.55 that of the Sun.
https://en.wikipedia.org/wiki/G-type_main-sequence_star#:~:text=A%20G%2Dtype%20main%2Dsequence,about%205%2C300%20and%206%2C000%20K.
A "typical" K0V type star should have luminosity of 0.46 that of the Sun.
https://en.wikipedia.org/wiki/K-type_main-sequence_star
And Sigma Draconis is believed to have a luminosity of 0.410 that of the Sun.
https://en.wikipedia.org/wiki/Sigma_Draconis
The square roots of those luminosities are 0.7416, 0.6782, and 0.6403, respectively. And if you decide on inner and outer limits of the Sun's circumstellar habitable zone, multiplying them by 0.7416, or 0.6782, or 0.6403 would give the limits of Sigma Draconis's circumstellar habitable zone according to whatever limits for the Sun's habitable zone you decide on.
What I call the Earth Equivalent Distance or EED is the distance from a star at which a planet would receive exactly as much radiation from the star as Earth gets from the Sun at the semi-major axis of Earth's orbit, which of course is 1 AU.
So multiplying 1 AU by 0.7416, 0.6782, or 0.6403 would give the distance from Sigma Draconis of its EED, where a planet would receive as much radiation as Earth gets from the Sun.
Fun fact: When I was a kid reading science fiction for the first time I had no knowledge of the Bayer designations of stars using Greek letters, and so Sigma Draconis was meaningless to me. And so I tended to think of Sigma Draconis as Stigma Draconis.