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A point moves such that the length of the tangent from it to the circle ${x^2} + {y^2} + 4x - 5y + 6 = 0$ is double the length of the tangent to the circle ${x^2} + {y^2} = 4$. Show that the locus is a circle. Find its centre and radius.

Answer Verified Verified
Hint: Here the tangent will be drawn from some point to two different circles. Length of the tangents will be computed and compared. This will give the equation which will be the locus of the point in general. This locus will be a circle. Further we will compute the center coordinate and radius of that circle.

Complete step-by-step answer:
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 $\dfrac{{\sqrt {305} }}{6}$.

Note: Coordinate geometry is used to represent various geometrical shapes through the coordinates and equation. These equations are representing the locus of general points lying on the given shape. Here the circle is the shape under consideration. We have used standard as well as general form of equations for the circle.