The locus of a point which is at a distance $r$ units from the given point $A (x_{1}, y_{1})$ is a circle which equation is $(x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}$

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Solution

Let the point be $(x, y)$

Distance from $(x_{1}, y_{1})$ is $r$ .

Using distance formula,

$\sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}} = r$

$⟹ (x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}$

Hence proved

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The locus of a point which is at a distance r units from the given point A(x1, y1) is a circle which equation is (x−x1)2+(y−y1)2=r2

Let the point be (x,y)Distance from (x1,y1) is r.Using distance formula,√(x−x1)2+(y−y1)2=r⟹(x−x1)2+(y−y1)2=r2Hence proved

The locus of a point which is at a distance $r$ units from the given point $A (x_{1}, y_{1})$ is a circle which equation is $(x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}$

Let the point be $(x, y)$

Distance from $(x_{1}, y_{1})$ is $r$ .

Using distance formula,

$\sqrt{(x - x_{1})^{2} + (y - y_{1})^{2}} = r$

$⟹ (x - x_{1})^{2} + (y - y_{1})^{2} = r^{2}$

Hence proved