Question

What does the equation ${(x-a)}^2+{(y-b)}^2=c^2$ become when it is transferred to parallel axes through
(1) the point $(a-c,b)$,
(2) the point $(a,b-c)$?

Answer 1

$(x-a{)}^2+(y-b{)}^2=c^2..........\left(1\right)$

Let a point on the circle with respect to old axis be $(x,y)$

Let the same point on the circle with respect to new axis be $(x^{\prime },y^{\prime })$

1) The axes are shifted by $(a-c,b)$

So, $x=x^{\prime }+a-c$ and $y=y^{\prime }+b$

Putting the value of $x$ and $y$ in equation $\left(1\right)$, we get

$(x^{\prime }+a-c-a{)}^2+(y^{\prime }+b-b{)}^2=c^2$

or, $(x^{\prime }-c{)}^2+y^{\prime 2}=c^2$

or, $x^{\prime 2}-2c{x}^{\prime }+c^2+y^{\prime 2}=c^2$

or, $x^{\prime 2}+y^{\prime 2}=2c{x}^{\prime }$

2) The axes are shifted by $(a,b-c)$

So, $x=x^{\prime }+a$ and $y=y^{\prime }+b-c$

Putting the value of $x$ and $y$ in equation $\left(1\right)$, we get

$(x^{\prime }+a-a{)}^2+(y^{\prime }+b-c-b{)}^2=c^2$

or, $x^{\prime 2}+(y^{\prime }-c{)}^2=c^2$

or, $x^{\prime 2}+y^{\prime 2}-2c{y}^{\prime }+c^2=c^2$

or, $x^{\prime 2}+y^{\prime 2}=2c{y}^{\prime }$