Question

Find the locus of $4{x^2} + 4{y^2} - 4x + 8y + 5 = 0$.

Answer 1

First of all, make the coefficients of ${x^2}$ and ${y^2}$ equal to one and then complete their whole squares by using simple math applications and algebraic identities. Then find the required locus.

Complete step-by-step answer:

Given equation is $4{x^2} + 4{y^2} - 4x + 8y + 5 = 0$

Dividing both sides with 4 we get

$$\Rightarrow \dfrac{{4{x^2} + 4{y^2} - 4x + 8y + 5}}{4} = \dfrac{0}{4} \\\ \Rightarrow \dfrac{{4{x^2}}}{4} + \dfrac{{4{y^2}}}{4} - \dfrac{{4x}}{4} + \dfrac{{8y}}{4} + \dfrac{5}{4} = 0 \\\ \Rightarrow {x^2} + {y^2} - x + 2y + \dfrac{5}{4} = 0 \\\$$

Grouping the same variables, we have

$\Rightarrow \left( {{x^2} - x} \right) + \left( {{y^2} + 2y} \right) + \dfrac{5}{4} = 0$

To complete the whole square, add and subtract ${\left( { - \dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$ in the first bracket and ${\left( {\dfrac{2}{2}} \right)^2} = 1$ in the second bracket

$$\Rightarrow \left( {{x^2} - x + \dfrac{1}{4} - \dfrac{1}{4}} \right) + \left( {{y^2} + 2y + 1 - 1} \right) + \dfrac{5}{4} = 0 \\\ \Rightarrow \left( {{x^2} - 2\left( {\dfrac{1}{2}} \right)\left( x \right) + \dfrac{1}{4}} \right) - \dfrac{1}{4} + \left( {{y^2} + 2\left( 1 \right)\left( y \right) + 1} \right) - 1 + \dfrac{5}{4} = 0 \\\ \Rightarrow \left( {{x^2} - 2\left( {\dfrac{1}{2}} \right)\left( x \right) + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) + \left( {{y^2} + 2\left( 1 \right)\left( y \right) + {1^2}} \right) = 1 + \dfrac{1}{4} - \dfrac{5}{4} \\\$$

We know that $\left( {{a^2} - 2ab + {b^2}} \right) = {\left( {a - b} \right)^2}$ and $\left( {{a^2} + 2ab + {b^2}} \right) = {\left( {a + b} \right)^2}$

$$\Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 1} \right)^2} = 1 - 1 \\\ \Rightarrow {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 1} \right)^2} = 0 \\\ \therefore S \equiv {\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 1} \right)^2} = 0 \\\$$

We know that for circle equation ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$ the centre is $\left( {h,k} \right)$ and radius is $r{\text{ cm}}$.

So, for the given circle the centre of the circle is $\left( {\dfrac{1}{2}, - 1} \right)$ and radius of the circle is $0$.

As the radius of the circle is zero, the given circle is a point circle at the point $\left( {\dfrac{1}{2}, - 1} \right)$.

Thus, the locus of the circle at $\left( {\dfrac{1}{2}, - 1} \right)$ with $0$ radius is given by ${\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + 1} \right)^2} = 0$ which is a point circle as shown in the below figure:

Note: A locus is a set of points that meet a given condition. The definition of circle locus of points a given distance from a given point in a two-dimensional plane. The given distance is the radius and the given point is the centre of the circle.