Question

Find the equation of the locus of the point which is at distance of 5 units from $(-2,3)$ 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 $d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ and find the distance of arbitrary point $(x,y)$ from given point $(-2,3)$. 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. $$$$
We know that the distance $d$ between any two points in plane with coordinates $(x_1,y_1),(x_2,y_2)$ is given by
$$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$$
We are asked in the question to find the equation of locus of the point which is at a distance of 5 units from $(-2,3)$ in a plane. $$$$

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