The eccentricity of the locus of point (3h + 2, k), where (h, k) lies on the circle x² + y² = 1, is

Let p = 3h + 2 and q = k
$\text{ or }\mathrm{h}=\frac{\mathrm{p}-2}{3}\text{ and }\mathrm{k}=\mathrm{q}$
Since (h, k) lies on x² + y² = 1, we have
h² + k² = 1
$\text{ or }{\left(\frac{\mathrm{p}-2}{3}\right)}^2+{\mathrm{q}}^2=1$
The locus is
${\left(\frac{\mathrm{x}-2}{3}\right)}^2+{\mathrm{y}}^2=1$
which has eccentricity
$\mathrm{e}=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}$