Check whether a line cuts the circle at real points.

Question

I am currently studying circles and I have come across a question in which I am doubtful that the given line even cuts the circle at real points.

If the line given to me is $$y=mx+c$$ and the equation of circle is $$x^2+y^2=r^2$$ then, is there a relationship between a,m and c through which I can check whether the line cuts the circle at two real points.

Answer 1

Points of intersection satisfy $$ x^2+(mx+c)^2=r^2$$

Which simplifies to $$(1+m^2)x^2+2mcx+c^2-r^2=0$$ The condition for having two real solutions is

$$4m^2c^2-4(1+m^2)(c^2-r^2)>0$$

After simplifying we get

$$(1+m^2)r^2>c^2$$

Answer 2

Hint: Solve the equation for $x$: $$x^2+(mx+c)^2=r^2$$ and distinguish three cases. I got this equation: $$x^2(1+m^2)+2mcx+c^2-r^2=0$$ Dividing by $1+m^2$ we get $$x^2+\frac{2mc}{1+m^2}x+\frac{c^2-r^2}{1+m^2}=0$$ Using the quadratic formual: $$x_{1,2}=-\frac{mc}{1+m^2}\pm\sqrt{\left(\frac{mc}{1+m^2}\right)^2-\frac{c^2-r^2}{1+m^2}}$$

Answer 3

Given line $L:y = mx + c$ and circle $C: x^2 + y^2 = r^2$ , on placing the value of $y$ in equation of $C$, you would obtain the following equation $$ x^2 ( 1 + m^2) + 2mxc + c^2 - r^2 =0 $$

Now, there are three possible cases, If the quadratic equation obtained has:

1. Two identical roots, then the given line $L$ is tangent to $C$.

2. Two distinct roots, then the given line $L$ intercepts the circle $C$.

3. Zero real roots, then the given line do not touch/intersect the circle.

Answer 4

If you draw a diagram you will note that the circle centred at the origin has radius $r$.

There are some lines which cut the circle and some which don't.

What distinguishes the cases is the distance from the origin to the nearest point on the line.

The line through the origin perpendicular to the original line has equation $x=-my$

You can proceed by computing the intersection of the two lines, and then checking whether the point of intersection is inside, on or outside the circle.

Obviously it is possible to compute directly and algebraically. This method is perhaps a little longer, but uses geometric insights.

Answer 5

Let $m\neq0$. The normal to $y=mx+c$ through the origin is given by the equation $y=-x/m$; determine the intersection point of both to find that the distance from $y=mx+c$ to the origin is $\frac{|c|}{\sqrt{1+m^2}}$. This is valid for $m=0$, too. No quadratics needed.