Let the center of the circle be $$(h, 2)$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$ \Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$ \Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$

The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$



Center $$(-1,-2)$$

Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$

then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$

$$ \Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$

Equation of circle whose center is $$\left( {h,k} \right)$$

i.e $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$



(radius of circle $$=k$$ because circle is tangent to $$x$$-axis)

Equation of circle passing through $$\left( { - 1, + 1} \right)$$

$$\therefore$$ $${\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}$$

$$ \Rightarrow 1 + {h^2} + 2h + 1 + {k^2} - 2k = {k^2}$$

$$ \Rightarrow {h^2} + 2h - 2k + 2 = 0$$

$$D \ge 0$$

$$\therefore$$ $${\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0$$

$$ \Rightarrow 1 + 2k - 2 \ge 0$$

$$ \Rightarrow k \ge {1 \over 2}$$

Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where

$$\angle AOB = {{2\pi } \over 3}$$



$$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$ \Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$ \Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$