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Let the given point P have the coordinate $({x_1},{y_1})$.

Now length of tangent from P to first circle ${C_1}$ with equation ${x^2} + {y^2} + 4x - 5y + 6 = 0$ will be,

$P{C_1}$=$\sqrt {{x_1}^2 + {y_1}^2 + 4{x_1} - 5{y_1} + 6} $

Similarly, length of tangent from P to second circle ${C_2}$ with equation ${x^2} + {y^2} = 4$will be,

$P{C_2}$=$\sqrt {{x_1}^2 + {y_1}^2 - 4} $

Now, we will compare the lengths of $P{C_1}$and double of$P{C_2}$as below,

$P{C_1} = 2 \times P{C_2}$

Squaring on both sides, we get

$P{C_1}^2 = 4 \times P{C_2}^2$

\[ \Rightarrow ({x_1}^2 + {y_1}^2 + 4{x_1} - 5{y_1} + 6) = 4({x_1}^2 + {y_1}^2 - 4)\]

$ \Rightarrow 3{x_1}^2 + 3{y_1}^2 - 4{x_1} + 5{y_1} - 22 = 0$

Thus the locus of all such points will be $3{x^2} + 3{y^2} - 4x + 5y - 22 = 0$

Hence it is proved that the above equation is the locus of all such points like $({x_1},{y_1})$ is a circle.

Now we will compute its center and radius.

Circle equation is $3{x^2} + 3{y^2} - 4x + 5y - 22 = 0$

We will do some algebraic manipulation to convert this equation to standard form.

$\Rightarrow {x^2} + {y^2} - \dfrac{4}{3}x + \dfrac{5}{3}y - \dfrac{{22}}{3} = 0 \\

\Rightarrow {x^2} - 2 \times \dfrac{2}{3}x + \dfrac{4}{9} + {y^2} + 2 \times \dfrac{5}{6}y + \dfrac{{25}}{{36}} - \dfrac{{22}}{3} - \dfrac{4}{9} - \dfrac{{25}}{{36}} = 0 \\

\Rightarrow {(x - \dfrac{2}{3})^2} + {(y - ( - \dfrac{5}{6}))^2} = {(\dfrac{{\sqrt {305} }}{6})^2} \\$

The above equation is the standard form of a circle.

Thus Center of the circle will be $(\dfrac{2}{3}, - \dfrac{5}{6})$ and radius $\dfrac{{\sqrt {305} }}{6}$.