Alevel Maths - Circles question?

Question

My teacher has set me questions for maths homework and I haven't even been taught how to do them. Here is one of the questions, if someone could do the question and explain what they are doing as they go along, that would be more than fantastic...

Question: Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.

Once I can do one, I can do the rest of them.

Thanks

- Luke :)

Answer 1

Find the coordinates of the points where the straight line meets 'y=1' meets the circle 'x^2+y^2-14x+2y+45=0'.

'x^2+1^2-14x+2(1)+45=0
x^2 -14x +48 = 0
the solutions are:
x=(14 +- sqrt( 14^2 -4(48) ) /2
x=14+- sqrt(4) )/2
x= (14 +2)/2 or x= (14-2)/2
x=8 or x = 6
.

Answer 2

This is perhaps the easiest type of these problems you could be given. The straight line y=1 is always going to be of the form (x, 1). So, just substitute 1 in the equation for the circle everywhere you see a y. Thus,

x^2 + (1^2) - 14x + (2*1) + 45 = 0
x^2 + 1 - 14x + 2 + 45 = 0
x^2 - 14x + 48 = 0
(x - 6)(x - 8) = 0
So, x = 6 or x = 8.

Thus, the two coordinates are (6, 1) and (8,1). Send me a message if you need more of an explanation.

Answer 3

substitute y = 1 to the equation x^2+y^2-14x+2y+45=0

so, x^2+1^2-14x+2(1)+45=0

x^2 +1 -14x+2+45=0

x^2 -14x+48=0

factorize it, you can get
(x - 8)(x - 6)=0, x = 6, 8

substitute x = 6 and x = 8, you will get four coordinates

(6, 1), (6, 3), (8, 1) and (8, 3)

Answer 4

Plug your y=1 into the eqn for the circle and solve for x. You'll get two values because the line intersects the circle at two locations. Note that once you do the substiution then you'll have a quadratic for x. Use the quadratic formula to solve, or factor, or complete the square, anyway you like.