SOLUTION: Find an equation of the line containing the centers of the two circles whose equations are given below. (x-6)^2+(y-3)^2=49 (x+4)^2+(y-4)^2=81

Question 1202856: Find an equation of the line containing the centers of the two circles whose equations are given below. (x-6)^2+(y-3)^2=49 (x+4)^2+(y-4)^2=81
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
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line containing the centers of the two circles whose equations are given below. (x-6)^2+(y-3)^2=49 (x+4)^2+(y-4)^2=81
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Say better this way:
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line containing the centers of the two circles whose equations are given below. (x-6)^2+(y-3)^2=49, AND (x+4)^2+(y-4)^2=81
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The two points are (6, 3) and (-4, 4).
IF you already know how to find equation of a line through two given points, then you're all set knowing what to do.

Do you understand why this is or can be used?

In more detail,
Picking the point (6, 3) as build the variable slope expression
and both points for the slope value

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Directly from the equations in the given form of the two circles, the centers are (6,3) and (-4,4).

The "run" from the first center to the second is -10 (from 6 to -4); the "rise" is 1 (from 3 to 4), so the slope (rise over run) is -1/10 = -0.1.

Use that slope in the general slope-intercept form of the equation using either of the two given points to find the equation.

ANSWER:

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