Question

Find the equation of the locus of the point which is at distance of 5 units from $\left( -2,3 \right)$ in a plane. \[\]

Answer 1

Hint: We recall the definition of locus. We use the distance formula for distance $d$ between any two points in plane with coordinates $d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ and find the distance of arbitrary point $\left( x,y \right)$ from given point $\left( -2,3 \right)$. We put $d=5$ as conditioned in the question and simplify after squaring both sides. \[\]

Complete step-by-step answer:
We know that locus is a shape formed by collection of points which satisfy a certain condition, for example the shape formed by collection at equal distance from end points of a line segment in a plane is called perpendicular bisector. \[\]
We know that the distance $d$ between any two points in plane with coordinates $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ is given by
\[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]
We are asked in the question to find the equation of locus of the point which is at a distance of 5 units from $\left( -2,3 \right)$ in a plane. \[\]

Let us assume that the coordinate of the arbitrary point be $\left( x,y \right)$. We use the distance formula between two points and find the distance of arbitrary point $\left( x,y \right)$ from given point $\left( -2,3 \right)$ as,
\[\begin{align}
  & d=\sqrt{{{\left( x-\left( -2 \right) \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
 & \Rightarrow d=\sqrt{{{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}} \\
\end{align}\]
We are given the question: the distance is always a fixed 5 units. SO we put $d=5$ in the above step and have
\[\Rightarrow 5=\sqrt{{{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}}\]
We square both side of the above step to have,
\[\Rightarrow 25={{\left( x+2 \right)}^{2}}+{{\left( y-3 \right)}^{2}}\]
Let us use the algebraic identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ for and proceed to have,
\[\begin{align}
  & \Rightarrow 25={{x}^{2}}+4x+4+{{y}^{2}}-6y+9 \\
 & \Rightarrow {{x}^{2}}+4x+{{y}^{2}}-6y-12=0 \\
\end{align}\]
The above equation is the required equation of locus. \[\]

Note: We note that the obtained locus is in the shape of a circle, in fact any locus of points which is at fixed distance from a fixed point will be the locus of a circle. The fixed point is the centre and the fixed distance is the radius of the circle. The equation of circle with centre $\left( a,b \right)$ and radius $r$ is given as ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$.