Find the equation of locus of a point which is equidistant from the coordinates axes . PLEASE ANSWER THE QUESTION
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The locus of point equidistant from a point is a circle
In the co-ordinate axes, the points are given by ordered pairs like (a, b)
The locus of points equidistant from (a, b) is given by the general equation of a circle:
(x - a)² + (y - b)² = r²
Where (a, b) is the centre of the circle, the point which the locus is equidistant to.
r = radius of the circle
In the co-ordinate axes, the points are given by ordered pairs like (a, b)
The locus of points equidistant from (a, b) is given by the general equation of a circle:
(x - a)² + (y - b)² = r²
Where (a, b) is the centre of the circle, the point which the locus is equidistant to.
r = radius of the circle
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Answer:
Step-by-step explanation:
If I understand correctly the locus of a point will be, all the possible points that satisfy your answer.
and that is, truly de definition of a circle that is centered on the coordinates axes.
that will mean, X^2+Y^2=R^2
the only tricky part here is that this circle has to truly pass for the point we have.
but we know that point will be (x’,y’)
based on that R^2 will be simply, x’^2 + y’^2
so your equation will finally be.
X^2+Y^2= x’^2 + y’^2 (where x’ and y’ are the coordinates of the point).
At least is what i could make up from your question
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