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I have 2 equations:

${x^2 + y^2 - 10x - 12y + 36 = 0}$

${x^2 + y^2 + 8x + 12y - 48 = 0}$

From this, the centre and radius of each circle is

(5, 6) and a radius of 5 (-4, -6) and a radius of 10

In order to prove that the circles touch externally the distance between the 2 centres is the same of the sum of the 2 radii or 15.

Using the distance formula I get

${{\sqrt {(-4 - 5)^2 + (-6 - 6)^2}}}$

Which is ${\sqrt {81 + 36} = \sqrt 225 = 15}$

So they touch externally but how can if I find the point where they intersect?

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  • $\begingroup$ sorry, I had a typo in the first equation. $\endgroup$ – dagda1 Sep 21 '15 at 20:34
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Hint:

The common point of the two circles is the intersection point of the straight line that passes through the two centers and one of the circles.

You can find it also solving the system between the equations of the circles (substitute one of the equations with the difference by the two).

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