Question 165840: find the equation of the circle touching the line x+2y=8 at (0,4)and passing through (3,7).
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! find the equation of the circle touching the line x+2y=8 at (0,4)and passing through (3,7).
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"touching" must mean tangent at that point, otherwise there's no single solution.
Since it's tangent, a diameter will be perpendicular to the line and pass thru (0,4). The slope of that line is +2, the negative inverse of the line x+2y=8.
The eqn of the line thru the center of the circle and thru (0,4) is:
y-4 = 2x
Now, the center of the circle is on that line and equidistant from (0,4) and (3,7).
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Any point (x,y) on line y = 2x+4 is, from (0,4) of sqrt[x^2 + (y-4)^2], and
from (3,7) sqrt[(y-7)^2 + (x-3)^2]. Note: (y-7) or (7-y) is the same, since it's squared.
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Set those equal:
sqrt[x^2 + (y-4)^2] = sqrt[(y-7)^2 + (x-3)^2]
x^2 + y^2-8y+16 = y^2-14y+49 + x^2-6x+9
-8y+16 = -14y+49 - 6x+9
6y = -6x + 42
y = -x + 7
This is the center, so it's on the line y = 2x+4
So -x + 7 = 2x+4
3x = 3
x = 1
y = 6
The center is (1,6)
To check:
(1 + 4) = 4 + 1 (squares of distances)
It is equidistant, and it's on the line, so it's correct.
Jeez
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