Contact points of circles?

Question

If 2 circles have equations:

Circle A => x^2 + y^2 -10x -12y +36

Circle B => x^2 + y^2 + 8x +12y -48

Find the point of contact that both of these circles meet at.

N.B. There is a question identical to this in my math textbook. I know the answer is (2,2), I just need to know how to find it i.e. a solution.

If it helps speeds things up.
the centres and radii are (5,6);5 and (-4,-6);10 respectively

I am still confused as to how you arrive at your solution of 25x^2 -100x + 100.

I have isolated my terms, but when I procede to multiply out, and sort the terms I am left with an unfactorisable mess

Answer 1

You may calculate the distance between the centers and compare it with the sum of radii. If the distance > sum, then there is no contact. If the distance=sum then there is only one contact point. If the distance

But If you did not make the test before, you will find the answer or if there is no answer. You must solve a system of equations.

1) subtract the two equation and you will find 3x + 4y -14 =0(after simplifying each term by -6)

2) Isolate y = (14 -3x)/4

3) Put the y in the first equation and evaluate all algebraic simplifications... finding 25x^2 -100x + 100 =0 or x^2 -4x+4=0


4) Try to solve the last equation: there is only a solution x=2.

5) Now put x=2 in the equation y =(14-3x)/4 and find y = 2

Answer 2

You forgot to put " = 0" after each of your "equations."

However, from the other information you provided (the radii in addition to the centres, the latter obvious from the x and y terms alone), it's easy to check that " = 0" is indeed all that's missing.

I'll let others tell you how to do it by algebra (it involves elimination, of course.) Here, however, is how you can do it using the lost art of geometric visualization.

Imagine the two centres put onto a graph. (You don't need actual graph paper; a simple sketch will do.) Now: what is the DISTANCE between these two centres? The first centre is at (9, 12) in separation with respect to the second centre. But this is 3 x (3,4). The latter dimensions are the x and y components of a Pythagorean triple!; so the missing length (the hypoteneuse), in other words the distance between the two centres, is 3 x 5 (the other number in the triple), = 15.

BINGO! With radii of 5 and 10 respectively, the circles DO have a point of contact. (5 + 10 = 15.)

Clearly, then, the contact point is just a third of the way from the first centre to the second. Therefore, it's at:

(5,6) - (9,12)/3 = (5 - 3, 6 - 4) = (2,2).

So you are correct! It IS the same problem as in your math textbook, and it has the same ultimate solution, though possibly not by this method.

Live long and prosper.

P.S. The algebraic method is tedious and clumsy; I just took a look at it:

Subtracting one eqn from the other, yes, you get rid of the squared terms. Unfortunately, you are still left with a linear relationship between x and y. You then have to solve for one of these variables in terms of the other (your choice), and then susbstitute back into one of the original eqns. If you're lucky and make no mistakes, this gruesome and asymmetric process should lead to an equation that will be a multiple of either (x - 2)^2 = 0 or (y - 2)^2 = 0 (since at a point of contact, there has to be a double root). Then plug that into any of the other eqns and find that the other variable is also 2.

This is a very nice example of a problem where a truly grungy algebraic approach will get you there in the end. However, despite the way math is taught in the U.S., there's far more to math than mere algebraic manipulation. Looking at things from another point of view (literally, in this case!) leads to a much more aesthetically satisfying solution; such alternative approaches should be encouraged more.

Answer 3

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