Find the locus of \[4{x^2} + 4{y^2} - 4x + 8y + 5 = 0\].
VerifiedHint: 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.
