Distance between points - equation of a line

Question

I have worked on this particular example: The distance between the point $M_1(3,2)$ and $A$ is $2\sqrt5$ and the distance between the point $M_2(-2, 2)$ and $B$ is $\sqrt5$. Come up with a equation for the line going through $A$ and $B$.

I have tried playing around with the equations for the circles with centers in $M_1$ and $M_2$ respectively (the radii being the distance between $M_1$ and $A$, that is the distance between $M_2$ and $B$ for the other circle), but I am stuck.

Would appreciate any help.

Thanks in advance.

P.S. I am probably wrong but, isn't there infinitely many solutions since with the data I'm given I am basically being asked to come up with a particular equation for a line connecting any two points on those circles ?

Answer 1

The way the question stands now, it has no answer. You do not know exactly where $A$ and $B$ are, as you only know they both lie on a circle. However, without knowing any other information, there are infinitely many pairs of points $A,B$ that satisfy your condition, and each will yield a different line.

Answer 2

Here is a diagram to your problem. Yes: there are infinitley many lines that go through both circles. In the diagram the red dot represents $M_2$ and the blue dot represents $M_1$

Answer 3

Hint:

As noted in the other answer your intuition that there are infinitely many solutions is correct. All these soution can be represented as a family of straight lines dependent on a parameter in this way:

1) Represent the points $P_1$ of the circle of center $O_1=(\alpha_1,\beta_1)$ and radius $r_1$ in parametric form as: $$ P_1=(\alpha_1 + r_1 \cos \theta, \beta_1+r_1 \sin \theta) $$ an do the same for the other circle of center $O_2=(\alpha_2,\beta_2)$ and radius $r_2$ $$ P_2=(\alpha_2 + r_2 \cos \theta, \beta_2+r_2 \sin \theta) $$ now the equations of the straight lines passing thorough $P_1$ and $P_2$ have the form: $$ y-(\alpha_1 +r_1\sin \theta)=\dfrac{\beta_1-\beta_2+(r_1-r_2)\sin \theta}{\alpha_1-\alpha_2+(r_1-r_2)\cos \theta} \left(x-(\alpha_1+r_1\cos \theta) \right) $$

Answer 4

Point on the circle with radius $2\sqrt{5}$ and center $(3,2)$ is at: $$ A = (2\sqrt{5} \cos(a) + 3, 2\sqrt{5} \sin(a) + 2)$$

The other circle's point is at:

$$ B = (\sqrt{5} \cos(b) - 2, \sqrt{5} \sin(b) + 2) $$

Then slope of the line is:

$ m = \frac{B_y - A_y}{B_x-A_x} = \frac { \sqrt{5} \sin(b) + 2 - (2\sqrt{5} \sin(a) + 2) }{ \sqrt{5} \cos(b) - 2 - (2\sqrt{5} \cos(a) + 3)} = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5}$

Then the equation of the line is: $$ y-A_y = m(x-A_x)$$ $$ y-(2\sqrt{5} \sin(a) + 2) = m(x-(2\sqrt{5} \cos(a) + 3)) $$ $$ y-(2\sqrt{5} \sin(a) + 2) = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5} (x-(2\sqrt{5} \cos(a)+3)) $$