If a circle passes through the point $$(a, b)$$ and cuts the circle $$x^2 + y^2 = 4$$ orthogonally, then the locus of its centre is

Let the general equation of the variable circle be:

$$ x^2 + y^2 + 2gx + 2fy + c = 0 $$

The coordinates of the centre of this general circle are given by:

$$ (-g, -f) $$

Let the coordinates of this centre be denoted as a moving point $$ (h, k) $$. This assumption gives the relations:

$$ h = -g $$

$$ k = -f $$

$$ g = -h $$

$$ f = -k $$

The problem states that the circle passes through the point (a, b). Substituting $$ x = a $$ and $$ y = b $$ into the general equation yields:

$$ a^2 + b^2 + 2ga + 2fb + c = 0 $$

Substitute the values of g and f in terms of h and k into this condition:

$$ a^2 + b^2 + 2(-h)a + 2(-k)b + c = 0 $$

$$ a^2 + b^2 - 2ah - 2bk + c = 0 $$

Now, consider the given fixed circle equation:

$$ x^2 + y^2 = 4 $$

$$ x^2 + y^2 - 4 = 0 $$

For this fixed circle, the coefficients are $$ g_1 = 0 $$, $$ f_1 = 0 $$, and the constant term is $$ c_1 = -4 $$.

The condition for two circles to cut each other orthogonally is:

$$ 2gg_1 + 2ff_1 = c + c_1 $$

Substitute the values of the fixed circle into this orthogonal condition:

$$ 2g(0) + 2f(0) = c + (-4) $$

$$ 0 = c - 4 $$

$$ c = 4 $$

Now, substitute this value of c back into the equation obtained from the passing point condition:

$$ a^2 + b^2 - 2ah - 2bk + 4 = 0 $$

Rearrange the terms to group the variable terms together:

$$ 2ah + 2bk - (a^2 + b^2 + 4) = 0 $$

To find the final equation for the locus of the center, replace the moving coordinates (h, k) with the general coordinate variables (x, y):

$$ 2ax + 2by - (a^2 + b^2 + 4) = 0 $$

Final Answer:

The locus of the centre of the circle is the line $$ 2ax + 2by - (a^2 + b^2 + 4) = 0 $$.