Basic circles question (x^2 + y^2 -8x -2y -3 = 0)

A circle with centre C has equation x²+y²-8x-2y-3=0.

i) The points S and T lie on the circumference of the circle. M is the midpoint of the chord ST. Given that the length of CM is 2, calculate the length of the chord ST.

ii) Find the coordinates of the point where the circle meets the line x-2y-12=0

Please note, this is a 'non-calculator' question.

Please explain how you found the answer to this question if possible (and show working). I'm not really sure how to approach answering it.

Thanks very much!!

The short answer:

i) ST = 8
ii) (6, -3)

The long answer:
As always, it helps to draw a picture of this kind of thing, to see what it is you are trying to solve.

I would start out by re-writing the equation for the circle. The normal approach is by completing the square:

x^2 + y^2 -8x -2y -3 = 0
(x - 4)^2 - 16 + (y - 1)^2 - 1 - 3 = 0
(x - 4)^2 + (y - 1)^2 = 20

In general, we have the formula:
(x - h)^2 + (y - k)^2 = r^2
for a circle with center (h, k) and radius r.

So in this case, we know the circle has center C = (4, 1) and radius sqrt(20) = 2 * sqrt(5)

Now, given the information in point i) of your problem, we can say that:
CMS (or equivalently, CMT) is a right triangle with CM = 2 and hypotenuse CS = 2 * sqrt(5)

By the pythagorean theorem, we know:
|CS|^2 = |CM|^2 + |MS|^2
20 = 4 + |MS|^2
16 = |MS|^2
MS = 4

Now, since M is the midpoint of ST, and MS = 4, then we have:

i) ST = 8

For part ii), we can rewrite the second equation in terms of y:
y = x/2 - 6

and plug this value in for y in the first equation:

x^2 + (x/2 - 6)^2 -8x -2(x/2 - 6) -3 = 0

simplifying gives:
x^2 + (x/2 - 6)^2 -8x -2(x/2 - 6) -3 = 0
x^2 + (x^2)/4 - 6x + 36 -8x -x + 12 - 3 = 0
(5/4)(x^2) - 15x + 45 = 0
(5/4)((x - 6)^2) = 0 (alternately, use the quadratic formula)