SOLUTION: Find an equation(s) of the circle(s) of radius 4 with center on the line 4x + 3y + 7 = 0 and tangent to 3x + 4y + 34 = 0

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Question 1068273: Find an equation(s) of the circle(s) of radius 4 with center on the line 4x + 3y + 7 = 0 and tangent to 3x + 4y + 34 = 0

Found 2 solutions by Alan3354, Fombitz:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Find an equation(s) of the circle(s) of radius 4 with center on the line 4x + 3y + 7 = 0 and tangent to 3x + 4y + 34 = 0
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There are 2 circles.
Find the equations of the 2 lines parallel to 3x + 4y + 34 = 0 and 4 units from it.
3x + 4y + 34 = 0
y = (-3/4)x - 17/2
Slope m of 3x + 4y + 34 = 0 is -3/4.
Difference in y-ints = 4/(cos(atan(-3/4)) = 5
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--> the 2 parallel lines are:
y = (-3/4)x - 17/2 + 5 = (-3/4)x - 7/2
and y = (-3/4)x - 17/2 - 5 = (-3/4)x - 27/2
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The intersection of those 2 lines and 4x + 3y + 7 = 0 are the 2 centers of the circles, (h,k).
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4x + 3y + 7 = 0
y = (-3/4)x - 7/2
4x -9x/4 - 21/2 = -7
7x/4 = 7/2
x = 2, y = -5
--> %28x-2%29%5E2+%2B+%28y%2B5%29%5E2+=+16
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Find the other circle the same way.
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email via the TY note for help or to check your work.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Find the line perpendicular to the tangent line that goes through the circle center.
3x%2B4y%2B34=0
4y=-3x-34
y=-%283%2F4%29x-17%2F2
Perpendicular lines have slopes that are negative reciprocals.
-%283%2F4%29m%5B2%5D=-1
m%5B2%5D=4%2F3
Let the circle center be located at (h,k) and let the intersection point of the tangent line and the perpendicular line be (a,b).
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So you know that,
1.3a%2B4b%2B34=0
2.4h%2B3k%2B7=0
The perpendicular line goes through both (h,k) and (a,b) so,
3.k-b=%284%2F3%29%28h-a%29
and from the distance formula,
4%5E2=%28h-a%29%5E2%2B%28k-b%29%5E2
4.%28h-a%29%5E2%2B%28k-b%29%5E2=16
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Substitute 3 into 4,
%28h-a%29%5E2%2B%2816%2F9%29%28h-a%29%5E2=16
%2825%2F9%29%28h-a%29%5E2=16
%28h-a%29%5E2=%2816%2A9%29%2F25
h-a=+0+%2B-+12%2F5
First case,
h=a+%2B+12%2F5
So then going back through the equations and substituting for h,
5. 3a%2B4b=-34
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4h%2B3k%2B7=0
4%28a%2B12%2F5%29%2B3k=-7
4a%2B3k=-7-12%2F5
6.4a%2B3k=-83%2F5
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k-b=%284%2F3%29%2812%2F5%29
7.k-b=16%2F5
You have three equations, three unknowns, using Cramer's rule,
a=-2%2F5
h=-41%2F5
k=-5
and then,
h=-2%2F5+%2B+12%2F5
h=10%2F5
h=2
So the first circle is,
highlight%28%28x-2%29%2B%28y%2B5%29%5E2=16%29
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Second case,
h=a+-+12%2F5
Again substituting as needed for h,
8. 3a%2B4b=-34
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4%28a-12%2F5%29%2B3k%2B7=0
4a%2B3k=-7%2B48%2F5
9.4a%2B3k=13%2F5
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k-b=%284%2F3%29%28-12%2F5%29
10.k-b=-16%2F5
Again you have three equations, three unknowns, using Cramer's rule,
a=754%2F35
h=-863%2F35
k=-195%2F7
and then,
h=a+-+12%2F5
h=754%2F35-12%2F5
h=134%2F7
So the second circle is,
highlight%28%28x-134%2F7%29%2B%28y%2B195%2F7%29%5E2=16%29
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