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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 ?

rogerl
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n4869
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  • I think you last guess is right. The question is asking you to "come up with 'a' equation ...". So, just return the most convenient one, the equation of the common chord. – Mick Nov 27 '15 at 16:44

4 Answers4

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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: $$ y-(\alpha_1 +r_1\sin \theta)=\dfrac{\beta_1-\beta_2+(r_1-r_2)\sin \theta}{\alpha_1-\alpha_2+(r_1-r_2)\cos \theta} \left(x-(\alpha_1+r_1\cos \theta) \right) $$

Emilio Novati
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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.

5xum
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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 $M_1$

Red dot represents $M_2$ and blue dot represents $M_1$

Rodrigo Zepeda
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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: $$ y-A_y = m(x-A_x)$$ $$ y-(2\sqrt{5} \sin(a) + 2) = m(x-(2\sqrt{5} \cos(a) + 3)) $$ $$ y-(2\sqrt{5} \sin(a) + 2) = \frac{\sqrt{5}(\sin(b)-2\sin(a))}{\sqrt{5}(\cos(b)-2\cos(a))-5} (x-(2\sqrt{5} \cos(a)+3)) $$

tp1
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