C(5,3) is the centre of a circle of radius 5 units. Show that this circle cuts the x-axis at A(1,0) and B(9,0)
im guessing simply drawing it with a compass is not what im being asked here. i dont know how to approach this
edit: im guessing pythorean lengths of the coordinates involved
Maybe you can use the circle relation since they are giving you coordinates
The relation for your circle is: \begin{align*} (x-5)^2 + (y-3)^2 & =25 \ \end{align*}
WHERE: (5, 3) is the center and radius is 5
Any point that is on the circle will satisfy this relation. If you plug x and y values of any point into the left hand side, if the result is 25 you have a point on your circle.
You can do this to A and B to check if they are on the circle
You are given everything you need to construct the equation of the circle.
If $(a, b)$ is the center of a circle, and $r$ is its radius, then the equation of the circle is given by $$(x-a)^2 + (y-b)^2 = r^2$$
In your case, since the center is given to be $(5, 3)$, and radius is $5$, our equation of the circle is given by $$(x - 5)^2 + (y -3)^2 = (5)^2 = 25.$$
Now that we have the equation of the circle, simply show show that $(1, 0)$, and $(9,0)$ satisfy the equation. You can, alternatively, solve for $x$ when $y = 0$, as any point on the $x$-axis has a $y$-coordinate equal to $0$. Solving for $x$ when $y = 0$ will give you two solutions for $x$: $x = 1$, and $x = 9$.
