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Recognising that $(x-a)^2+(y-b)^2=r^2$ is a circle of radius $r$ with centre $(a,b)$

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You are given the equation $\\simplify{(x-{a})^2+(y-{b})^2={rs}}$.

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The equation $\\simplify{(x-{a})^2+(y-{b})^2={rs}}$ is the equation of a circle with radius [[0]] centred at the point $\\large($ [[1]], [[2]]$\\large)$.

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The equation $(x-x_0)^2+(y-y_0)^2=r^2$ is the equation of a circle centred at the point $(x_0,y_0)$ with radius $r$.

Therefore $\\simplify{(x-{a})^2+(y-{b})^2={rs}}$ represents a circle with centre $(\\var{a}, \\var{b})$ and radius $\\var{r}$ since we can write the equation as $\\simplify[!basic]{(x-{a})^2+(y-{b})^2={r}^2}$.

The above information allows us to find four points of the circle easily:

The top of the circle is at the point $\\large($[[0]], [[1]]$\\large)$
The bottom of the circle is at the point $\\large($[[2]], [[3]]$\\large)$
The far left of the circle is at the point $\\large($[[4]], [[5]]$\\large)$
The far right of the circle is at the point $\\large($[[6]], [[7]]$\\large)$

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Since we know the centre of the circle is $(\\var{a}, \\var{b})$ and we know the radius is $\\var{r}$ we can find four points of the circle easily (since a circle is the set of all points one radius distance from the centre):

The top of the circle is $\\var{r}$ spaces above the centre, that is, at the point $(\\var{a},\\var{b}+\\var{r})=(\\var{a},\\var{b+r})$
The bottom of the circle is $\\var{r}$ spaces below the centre, that is, at the point $(\\var{a},\\var{b}-\\var{r})=(\\var{a},\\var{b-r})$
The far left of the circle is $\\var{r}$ spaces to the left of the centre, that is, at the point $(\\var{a}-\\var{r},\\var{b})=(\\var{a-r},\\var{b})$
The far right of the circle is $\\var{r}$ spaces to the right of the centre, that is, at the point $(\\var{a}+\\var{r},\\var{b})=(\\var{a+r},\\var{b})$

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