I'm very stuck... Circles

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

Find the shortest distance between the line and origin. That will be the maximum radius of the circle

The point on the line closest to the origin is its point of intersection with a line perpendicular to it, and through the origin.

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

2/3???

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.

$y=3x-2$

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

Correct. So, the line perpendicular to $y=3x+2$ (and through the origin) is $y=-\frac{x}{3}$

Now find the point of intersection of those lines.

3(3x+2)=-x
9x +6=-x
10x+6=0
2(5x+3)=0
5x+3=0
X=-3/5

I don't follow this.


Sorry - I misquoted the first line - it's $3x-2$ (I'd put +), which gives $x=+\frac{3}{5}$ and $y=\frac{-1}{5}$.

By Pythagoras, we get $d=\sqrt{(\frac{3}{5})^2+(\frac{1}{5})^2}=\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}}$

I don't follow this.


Sorry - I misquoted the first line - it's $3x-2$ (I'd put +), which gives $x=+\frac{3}{5}$ and $y=\frac{-1}{5}$.

By Pythagoras, we get $d=\sqrt{(\frac{3}{5})^2+(\frac{1}{5})^2}=\sqrt{\frac{10}{25}}= \sqrt{\frac{2}{5}}$

How did you get 3/5 and 1/5?

The first line is $y=3x-2$

The product of the gradients of perpendicular lines is -1. The line perpendicular to this, and through the origin, is therefore $y=\frac{-x}{3}$

These lines intersect when:

$\frac{-x}{3}=3x-2$

$\therefore \frac{10x}{3}=2$

$\therefore x=\frac{3}{5}$

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

$\therefore y=\frac{-x}{3}=\frac{-\frac{3}{5}}{3}=\frac{-3}{15}=\frac{-1}{5}$

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 $\frac{-1}{5}$, I dropped the minus sign, as it doesn't make any difference.