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Question 723076: equation of the circle that passes through (9,7) and is tangent to both y-axis and the line 3x-4y-24=0.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! equation of the circle that passes through (9,7) and is tangent to both y-axis and the line 3x-4y-24=0.
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The line,3x-4y-24=0, has a y-intercept at (0,-6) and a x-intercept at (8,0)
This makes a right triangle with hypotenuse=10 which is=distance from the y-intercept (0,-6) to the tangent point(0,4) the circle makes with the y-axis.
The y-coordinate(k) of the center=4.
We now have two equations of the circle we can work with.
(9-h)^2+(7-k)^2=r^2 (using given point (9,7))
(0-h)^2+(4-k)^2=r^2 (using tangent point (0,4))
k=4
..
81-18h+h^2+(7-4)^2=r^2
h^2+(4-4)^2=r^2
..
81-18h+h^2+9=r^2
h^2+(0)^2=r^2
subtract
18h=90
h=5
center of circle: (h,k)=(5,4)
solving for radius using coordinates of center and given point (9,7)
(9-h)^2+(7-k)^2=4^2+3^2=16+9=25
Equation of given circle: (x-5)^2+(y-4)^2=25
The graph below will give you some idea what the configuration looks like
y=(25-(x-5)^2)^.5+4
y=3x/4-6
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