The line 3x + 4y = 24 cuts the x axis at point A and the y axis at point B. The point C is the mid-point of AB.

(i) Find the co-ordinates of A, B and C. Also, find the length of AB.
(ii) Find the equation of the circle with centre C which passes through A and B. Show that this circle also passes through the origin.
(iii) The tangent to the circle at B meets the straight line through 0 and C at the point D. Find the co-ordinates of D.

Can you also tell me how to work this out please?

Thank you, very much appreciated.

Okay, to find A, or where the line cuts the x-axis (x-intercept), set y = 0 in the line equation:

3x + 4*0 = 24
3x = 24
x = 8
You have the x-intercept coordinate: A = (8,0)

To find B, or where the line cuts the y-axis (y-intercept), set x = 0 in the line equation:

3*0 + 4y = 24
4y = 24
y = 6
You have the y-intercept coordinate: B = (0,6)

To find the midpoint C, use the formula:
((x1 + x2) / 2 , (y1 + y2) / 2)

From the two points above, it would be:
((8+0)/2 , (0+6)/2)
C = (4,3)

To find the distance, use this formula:

d = sqrt((0-8)^2 + (6-0)^2)
d = sqrt(64 + 36)
d = sqrt(100)
d = 10 units

(ii) The equation used is the standard equation that has the form

(x - h)2 + (y - k)2 = r2
where h and k are the x- and y-coordinates of the center of the circle and r is the radius.

So, in using the point C = (4,3), and since the distance, which is also the diameter of the circle, we can find this equation. Note, we need the radius, so divide the distance by 2, to get 5:

We now have the form for the circle:
(x-4)^2 + (y-3)^2 = 25

I'm guessing to show that the circle passes through the origin, you would substitute x = 0 and y = 0 into the circle equation and make sure it checks out:

(0-4)^2 + (0-3)^2 = 25
16 + 9 = 25
25 = 25

When you say meets the straight line through 0 and C at the point D. is the 0 meaning x = 0 or y = 0

To be honest I'm not sure. It could mean the origin? If you don't know what it means, that's fine. Thanks for all your help.