Find the values of b for which the line $y=3x+b$ intersects the circle $x^{2}+y^{2}=4$ I'm not sure how to tackle this question. First thought was to sketch both the line and the circle together, but it doesn't make it much clearer how to find b. Solving for both equations for y and equating them also arrives to the wrong coclusion.
Edit: The correct answer to this question is $b \leq 2\sqrt{10}$. The whole thing is arriving to this result and actually understanding it.
 A: Hint: Plug $y=3x+b$ into the other equation. You will get a quadratic equation with respect to $x$. When the two graphs intersect, the quadratic equation would have real solutions.
A: Geometrically, you want to find the value of $3a$ in the following picture:

Now, using similar triangles and the Pythagorean Theorem, we have
$$3=\frac{2}{\sqrt{a^2-4}}\Longrightarrow 3a=\pm 2\sqrt{10}.$$
The value of $b$ therefore must be in the interval $[-2\sqrt{10},2\sqrt{10}].$
A: To find an intersection point, we can substitute one equation into the other; in this case we get
$$ 4 = x^2 + (3x+b)^2 = 10x^2 + 6bx + b^2 .$$
This is both an equation in $x$ and in $b$, so we could use quadratic formula to solve for either. The "incorrect" values of $b$ will occur when there is a negative square root in quadratic formula, which is probably easier to see if we solve for $x$ (because then the $b$ will be under the root):
$$ x = \frac{-6b\pm\sqrt{36b^2-40(b^2-4)}}{20} $$
so we need $36b^2-40b^2+160 = -4b^2 + 160$ to be non-negative, ie $|b| \leq \sqrt{40} = 2\sqrt{10}$.
