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# I'm very stuck... Circles watch

1. the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
So far I've done substituted 0 for x and then for y. Whereby: (0,-2) and (2/3,0)
I've used the formula (x-a)^2 +(y-b)^2=r^2
(0-2/3)^2+ (-2,0) ^2 =r^2
4/9+4=r^2
40/9=r^2
2 root 10 divided by 3
But this is definitely not correct
Where have I gone wrong

Thanks
2. Set up an equation as if you were trying solve the simultaneous equations. Then you have a quadratic which I will have no solutions if the curves don't cross - think discriminant.
3. Find the shortest distance between the line and origin. That will be the maximum radius of the circle
4. (Original post by Musicanor)
the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
I'd tackle this slightly differently (I do that a lot).

The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.
5. (Original post by Laissez‒faire)
Find the shortest distance between the line and origin. That will be the maximum radius of the circle
2/3???
6. (Original post by RogerOxon)
The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.

What is the equation of a line perpendicular to that, through the origin? (Hint: the product of the gradients of two perpendicular lines is -1)
7. (Original post by Musicanor)
the line with equation y=3x-2 does not intersect the circle with centre (0,0) and radius r . find the range of possible values of r. Answer is 0<r< root 2/5
So far I've done substituted 0 for x and then for y. Whereby: (0,-2) and (2/3,0)
I've used the formula (x-a)^2 +(y-b)^2=r^2
(0-2/3)^2+ (-2,0) ^2 =r^2
4/9+4=r^2
40/9=r^2
2 root 10 divided by 3
But this is definitely not correct
Where have I gone wrong

Thanks
Take the line perpendicular to through the origin. What is the point of intersection between these two lines? Hence what is the distance between the origin and this point?

Get it?
8. (Original post by Musicanor)
2/3???
I haven't done it. You need to find an eq of a line perpendicular which goes through 0,0 and then find their intersection. Then find the distance between that and the origin and you have the maximum radius.
9. (Original post by RogerOxon)
I'd tackle this slightly differently (I do that a lot).

The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.
Thanks
I've got:
2/3--> -3/2
Y=3(0) -3/2
(0, -3/2)
0=3x-3/2
3/2= 3x
3/2 / 3=X
1/2=x

(0-1/2)^2 + (-3/2 -0)^2=r^2
1/4 +9/4 =r^2
5/2=r^2
Root 5/2 =r

But answer is meant to be root 2/5
???
10. (Original post by RogerOxon)

What is the equation of a line perpendicular to that, through the origin? (Hint: the product of the gradients of two perpendicular lines is -1)
-1/3
11. (Original post by Musicanor)
-1/3
Correct. So, the line perpendicular to (and through the origin) is

Now find the point of intersection of those lines.
12. (Original post by RogerOxon)
Correct. So, the line perpendicular to (and through the origin) is

Now find the point of intersection of those lines.
Y=3x+2
Y=-x/3
Y-y1= m(x-X1)
Y-2= -x/3 (x-0)
Y-2= -x/3 x
Y= -x/3 +2

3(3x+2)=-x
9x +6=-x
10x+6=0
2(5x+3)=0
5x+3=0
X=-3/5
Is this close:
2/3--> -3/2 Y=3(0) -3/2 (0, -3/2) 0=3x-3/2 3/2= 3x 3/2 / 3=X 1/2=x (0-1/2)^2 + (-3/2 -0)^2=r^2 1/4 +9/4 =r^2 5/2=r^2 Root 5/2 =r
Thank you so much for helping me
13. (Original post by Musicanor)
Y-y1= m(x-X1)
Y-2= -x/3 (x-0)
Y-2= -x/3 x
Y= -x/3 +2

3(3x+2)=-x
9x +6=-x
10x+6=0
2(5x+3)=0
5x+3=0
X=-3/5
Sorry - I misquoted the first line - it's (I'd put +), which gives and .

By Pythagoras, we get
14. (Original post by RogerOxon)

Sorry - I misquoted the first line - it's (I'd put +), which gives and .

By Pythagoras, we get
Yay Thanks I understand now
15. (Original post by RogerOxon)

Sorry - I misquoted the first line - it's (I'd put +), which gives and .

By Pythagoras, we get
How did you get 3/5 and 1/5?
16. (Original post by Musicanor)
How did you get 3/5 and 1/5?
The first line is

The product of the gradients of perpendicular lines is -1. The line perpendicular to this, and through the origin, is therefore

These lines intersect when:

We can get the y coordinate by using that x value in the equation of either line:

The distance from the origin to the point of intersection i the sqare roto of the sum of the squares of the difference in x and the difference in y - from the origin, it's just the squares of the coordinates. When I squared the , I dropped the minus sign, as it doesn't make any difference.
17. (Original post by RogerOxon)
The first line is

The product of the gradients of perpendicular lines is -1. The line perpendicular to this, and through the origin, is therefore

These lines intersect when:

We can get the y coordinate by using that x value in the equation of either line:

The distance from the origin to the point of intersection i the sqare roto of the sum of the squares of the difference in x and the difference in y - from the origin, it's just the squares of the coordinates. When I squared the , I dropped the minus sign, as it doesn't make any difference.
Thank you so so so sooooo much. You've helped with a lot you're the best

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