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# Core 1 - Circles watch

1. A circle with the centre C has the equation (x-2) squared + (y+5) squared = 25.
Prove that the line 2y=x does not meet the circle.
2. (Original post by Chelsea12345)
A circle with the centre C has the equation (x-2) squared + (y+5) squared = 25.
Prove that the line 2y=x does not meet the circle.
What have you tried already?
3. Everywhere you see an x in the equation of the circle, write 2y instead (i.e. you have to try to solve the equations simultaneously).

Now expand the equation of the circle, simplify, and try to solve. What do you find?
4. (Original post by Chelsea12345)
A circle with the centre C has the equation (x-2) squared + (y+5) squared = 25.
Prove that the line 2y=x does not meet the circle.
What are your thoughts?

Hypothetically, if it does meet the circle, how would you find the points where they would meet?
5. (Original post by SeanFM)
What are your thoughts?

Hypothetically, if it does meet the circle, how would you find the points where they would meet?
If it does meets the circle, i would substitute the x value and y values in?
6. (Original post by KaylaB)
What have you tried already?
tbh, i havent thought of anything so far to work it out
7. (Original post by Chelsea12345)
If it does meets the circle, i would substitute the x value and y values in?
Correct. Does that help you on how to prove that it never meets the circle?
8. (Original post by Pangol)
Everywhere you see an x in the equation of the circle, write 2y instead (i.e. you have to try to solve the equations simultaneously).

Now expand the equation of the circle, simplify, and try to solve. What do you find?
i understand that you have to solve simultaneously to see if it crosses the the line 2y=x. but why would you put 2y intsead of x?
9. (Original post by Chelsea12345)
tbh, i havent thought of anything so far to work it out
Substitute values for y or x in, use the discriminant see if there are any real roots (i.e. b^2 - 4ac >= 0)
10. (Original post by Chelsea12345)
i understand that you have to solve simultaneously to see if it crosses the the line 2y=x. but why would you put 2y intsead of x?
Because you give the equation of the line as 2y = x. So, everywhere on the line, the value of x is the same as the value of 2y. If the circle and the line meet, it must be at a point where x = 2y. So put x = 2y in the equation of the circle, and see what happens.

How would you solve the equations simultaneously without doing this?
11. (Original post by Pangol)
Because you give the equation of the line as 2y = x. So, everywhere on the line, the value of x is the same as the value of 2y. If the circle and the line meet, it must be at a point where x = 2y. So put x = 2y in the equation of the circle, and see what happens.

How would you solve the equations simultaneously without doing this?
oh right.

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