Question

Find the locus of a point which remains at a distance of 2.2 centimetre from a fixed point

Answer 1

Answer:

$\large\boxed{\sf{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1}) }^{2} = {(2.2)}^{2} }}$

Step-by-step explanation:

Let the fixed point is (a,b)

And , the variable point is (c,d)

Also, it's given that the distance between these points is 2.2 cm

Therefore, by distance formula, we get

$= > \sqrt{ {(c - a)}^{2} + {(d - b)}^{2} } = 2.2 \\ \\ = > {(c - a)}^{2} + {(d - b)}^{2} = {(2.2)}^{2}$

Replacing the variables by x and y,

We get the required locus as

$\bold{= > {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1}) }^{2} = {(2.2)}^{2} }$

Clearly, it's an equation of a Circle