Question 1067101
duplicate.
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Show or email what you've tried.
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Here's a similar problem:
The point(3,4) is on the circle that is tangent to the line 2y-3x + 7 = 0 at (1,-6). Find the equation of the circle. Where does it cut the x-axis?
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Label the points.
The point A(3,4) is on the circle that is tangent to the line 2y-3x + 7 = 0 at B(1,-6). Find the equation of the circle. Where does it cut the x-axis?
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Find the perpendicular bisector of AB.  The center of the circle is on that line.
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A(3,4) and B(1,-6)
Midpoint is (2,-1)
The slope of AB is (-6-4)/(1-3) = 5
--> line thru the MP with slope 5 is y+1 = 5*(x-2) = 5x-10
y = 5x - 11
zat make sense?
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Now find the equation of the line perpendicular to the given line thru the tangent point (1,-5).  The center of the circle is also on that line.
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The slope of 2y-3x + 7 = 0 is 3/2
The slope of lines perpendicular is -2/3
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Slope of -2/3 thru (1,-6) --> y+6 = (-2/3)*(x-1) = -2x/3 + 2/3
y = -2x/3 + 20/3
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The center is the intersection of the 2 lines.
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y = -2x/3 + 20/3
y = 5x - 11
y = -2x/3 + 20/3 = 5x - 11
-2x/3 + 20/3 = 5x - 11
-2x + 20 = 15x - 33
17x = 53
x = 53/17
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y = 5x - 11
y = 78/17
--> the center of the circle is (53/17,78/17)
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The radius is the distance from the center to either point.
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{{{r^2 = diffx^2 + diffy^2)}}}
using the point (3,4)
{{{r^2 = (3 - 53/17)^2 + (4 - 78/17)^2)}}}
{{{r^2 = 4/289 + 100/289 = 104/289}}}
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The circle is {{{(x - 53/17)^2 + (y-78/17)^2 = sqrt(104)/17}}}
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That's a lot of work.
Can you show some work where you tried?
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Just getting the answer and moving on guarantees you won't know what you're doing down the road.