Posted by **Connie** on Friday, April 25, 2008 at 2:54pm.

two circles whose equations are (x-3)^2+(y-5)^2=25 and (x-7)^2+(y-5)^2=9 intersect in two points. What is the equation of the line passing through these two points?

Can anyone please give me some ideas how to do it without graphing it out? THANKS A LOT!

- MATH -
**Damon**, Friday, April 25, 2008 at 3:32pm
The centers are at (3,5) and at (7,5) along the horizontal line y = 5 and 7-3 = 4 apart

left radius = 5

right radius = 3

so

the points lie on vertical line x = something an equal distance above and below y = 5

so

triangle

base = 4 along y = 5 line

left leg = 5

right leg = 3

let's use law of cosines to find the angle on the left, between the y = 5 line and the long side length 5

3^2 = 5^2 + 4^2 - 2*5*4 cos A

9 = 25 + 16 - 40 cos A

32 = 40 cos A

A = 36.87 degrees

now the distance from x = 3 to our desired vertical line is

5 cos A which is 5 (32/40) = 4

so our line is

x = 3 + 4 = 7

which we could have said earlier seeing that our triangle is

3, 4, 5 and therefore a right triangle

- MATH -
**Jordan**, Friday, April 25, 2008 at 9:32pm
What do you mean?

- MATH -
**Jordan**, Friday, April 25, 2008 at 9:32pm
What do youm mean?

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