Equation of circle with center $$(0,3)$$ and radius $$2$$ is

$${x^2} + {\left( {y - 3} \right)^2} = 4$$

Let locus of the variable circle is $$\left( {\alpha ,\beta } \right)$$

As it touches $$x$$-axis.

$$\therefore$$ It's equation is $${\left( {x - \alpha } \right)^2} + {\left( {y + \beta } \right)^2} = {\beta ^2}$$



Circle touch externally $$ \Rightarrow {c_1}{c_2} = {r_1} + {r_2}$$

$$\therefore$$ $$\sqrt {{\alpha ^2} + {{\left( {\beta - 3} \right)}^2}} = 2 + \beta $$

$${\alpha ^2} + {\left( {\beta - 3} \right)^2} = {\beta ^2} + 4 + 4\beta $$

$$ \Rightarrow {\alpha ^2} = 10\left( {\beta - 1/2} \right)$$

$$\therefore$$ Locus is $${x^2} = 10\left( {y - {1 \over 2}} \right)$$ which is parabola.

Solving $$y=x$$ and the circle

$${x^2} + {y^2} - 2x = 0,$$ we get

$$x = 0,y = 0$$ and $$x=1,$$ $$y=1$$

$$\therefore$$ Extremities of diameter of the required circle are

$$\left( {0,0} \right)$$ and $$\left( {1,1} \right)$$. Hence, the equation of circle is

$$\left( {x - 0} \right)\left( {x - 1} \right) + \left( {y - 0} \right)\left( {y - 1} \right) = 0$$

$$ \Rightarrow {x^2} + {y^2} - x - y = 0$$

Let the variable circle be

$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ $${p^2} + {q^2} + 2gp + 2fq + c = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Circle $$(1)$$ touches $$x$$-axis,

$$\therefore$$ $${g^2} - c = 0 \Rightarrow c = {g^2}.\,\,\,$$

From $$(2)$$

$${p^2} + {q^2} + 2gp + 2fq + {g^2} = 0\,\,\,\,\,\,\,\,\,...\left( 3 \right)$$

Let the other end of diameter through $$(p,q)$$ be $$(h,k),$$

then, $${{h + p} \over 2} = - g$$ and $${{k + q} \over 2} = - f$$

Put in $$(3)$$

$${p^2} + {q^2} + 2p\left( { - {{h + p} \over 2}} \right) + 2q\left( { - {{k + q} \over 2}} \right) + {\left( {{{h + p} \over 2}} \right)^2} = 0$$

$$ \Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0$$

$$\therefore$$ locus of $$\left( {h,k} \right)$$ is $${x^2} + {p^2} - 2xp - 4yq = 0$$

$$ \Rightarrow {\left( {x - p} \right)^2} = 4qy$$

Let the variable circle is

$${x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)$$

It passes through $$(a,b)$$

$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$

$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally

$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$

$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$

$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is

$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$

or $$2ax + 2by = {a^2} + {b^2} + 4$$