The two circles $$x^2 + y^2 = ax$$ and $$x^2 + y^2 = c^2 \, (c > 0)$$ touch each other if:

The first circle is $$x^2 + y^2 = ax$$. Bring the $$x$$-terms together and complete the square:

$$x^2 - ax + y^2 = 0 \;\;\Rightarrow\;\; (x - \tfrac{a}{2})^2 + y^2 = (\tfrac{a}{2})^2.$$

Hence
  centre $$C_1\,( \tfrac{a}{2},\,0 )$$ and radius $$r_1 = \tfrac{|a|}{2}.$$

The second circle is $$x^2 + y^2 = c^2 \;(c \gt 0)$$ with
  centre $$C_2\,(0,\,0)$$ and radius $$r_2 = c.$$

Distance between centres:

$$d = |C_1C_2| = \sqrt{\left(\tfrac{a}{2}\right)^2 + 0^2} = \tfrac{|a|}{2}.$$

For two circles to touch, the distance between their centres equals the absolute difference (internal touch) or the sum (external touch) of their radii.

External touch would require $$d = r_1 + r_2,$$ i.e. $$\tfrac{|a|}{2} = \tfrac{|a|}{2} + c,$$ which forces $$c = 0,$$ impossible because $$c \gt 0.$$ Therefore only internal touching is possible.

Internal touch condition: $$d = |\,r_2 - r_1\,|.$$
So $$\tfrac{|a|}{2} = |\,c - \tfrac{|a|}{2}\,|.$$

Hence only Case 1 is valid, giving the condition $$|a| = c.$$

Option A which is: $$|a| = c$$