If a circle passes through the point $$(a, b)$$ and cuts the circle $$x^2 + y^2 = p^2$$ orthogonally, then the equation of the locus of its centre is

Let the required circle have centre $$(h,k)$$ and radius $$r$$.
Its equation is $$x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$$.

1. Condition “passes through $$(a,b)$$”
  Substitute $$(a,b)$$: $$a^2 + b^2 - 2ha - 2kb + (h^2 + k^2 - r^2) = 0$$.
  Hence $$r^2 = (h-a)^2 + (k-b)^2$$ $$-(1)$$.

2. Condition “cuts $$x^2 + y^2 = p^2$$ orthogonally”
  For two circles $$S_1: x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2)=0$$ and $$S_2: x^2 + y^2 - p^2 = 0$$,
  the orthogonality condition is $$2g g' + 2f f' = c + c'$$, where $$g=-h,\, f=-k,\, c=(h^2 + k^2 - r^2)$$ and $$g'=0,\, f'=0,\, c'=-p^2$$.
  Thus $$0 + 0 = (h^2 + k^2 - r^2) - p^2$$ giving $$h^2 + k^2 = r^2 + p^2$$ $$-(2)$$.

3. Eliminate $$r^2$$ using $$(1)$$ and $$(2)$$:
  $$h^2 + k^2 = (h-a)^2 + (k-b)^2 + p^2$$.

Expand the right side:
  $$h^2 + k^2 = h^2 - 2ah + a^2 + k^2 - 2bk + b^2 + p^2$$.

Cancel $$h^2 + k^2$$ on both sides and rearrange:
  $$2ah + 2bk = a^2 + b^2 + p^2$$.

4. Replace $$(h,k)$$ by the general point $$(x,y)$$ to obtain the locus:
  $$2ax + 2by - (a^2 + b^2 + p^2) = 0$$.

Therefore the correct option is
Option D which is: $$2ax + 2by - (a^2 + b^2 + p^2) = 0$$.