# (a) (i) A circle touches the x-axis and has its centre at the point (4,2). Write down the equation of this circle. (ii) Confirm that the point (3 , 2 + 3) is on the circle

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The formula for a circle is (x-h)^2 + (y-k)^2 = r^2

(h,k) is the center of the circle.

In this example, touching the x-axis also means the x axis is tangent to the circle, where y = 0. This also means it is perpendicular to the radius. And, the perpendicular/radius runs through the center of the circle, aka where x = 4. Since y = 0 at this point, it touches at (4,0) as well as r = 2. So, the formula would be

(x-4)^2 + (y-2)^2 = 2^2

(x-4)^2 + (y-2)^2 = 4

As the graph also shows:

To prove (3,5) is on the circle, plug the numbers in to assure equality:

(3-4)^2 + (5-2)^2 = 2^2

(-1)^2 + (3)^2 = 4

1 + 9 = 4

10 = 4

There is no equality, so (3,5) isn't on the line.

Now, if that's suppose to be (3, 2 + sqrt(3)), then, we would have:` `

(3-4)^2 + (2 + sqrt(3)-2)^2 = 2^2

(-1)^2 + (sqrt(3))^2 = 4

1 + 3 = 4

4 = 4

There is equality, so (3,2+sqrt3) is on the line.