SOLUTION: Find the equation of a circle whose center is on 2x+y=1, tangent to 3x-4y=10 and radius 5. thank you

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Question 1130502: Find the equation of a circle whose center is on 2x+y=1, tangent to 3x-4y=10 and radius 5. thank you
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-h%29%5E2%2B%28y-k%29%5E2=25
find intersection of line 2x%2By=1 and tangent 3x-4y=10
2x%2By=1.......solve for y
3x-4y=10
---------------------
y=1-2x
substitute in 3x-4y=10
3x-4%28-2x%2B1%29=10
3x%2B8x-4=10
11x=14
x+=+14%2F11
then,
y=1-2%2814%2F11%29
y+=+-17%2F11
intersection point is:
P( 14%2F11,+-17%2F11)

-> since center is on a line 2x%2By=1 , we have
2h%2Bk=1
=>k+=+1+-+2+h
so, center C (h,k)=(h,1+-+2+h)
distance from P and C is+r=5
5=sqrt%28%28h-14%2F11%29%5E2%2B%281-2h%2B17%2F11%29%5E2%29
h+=+14%2F11+-+sqrt%285%29
h=-0.96334+
or
h+=+14%2F11+%2B+sqrt%285%29
h=3.50879525
=>k+=+1+-+2+%28-0.96334%29
k=2.92668+
or
k=1+-+2+%283.50879525%29
k=-6.01759

so, we have two centers that lie on given line:
center C (-0.96,2.93 )
and
center C (3.5,-6)
then, we have two circles:
%28x%2B0.96%29%5E2%2B%28y-2.93+%29%5E2=25
or
%28x-3.5%29%5E2%2B%28y%2B+6%29%5E2=25






Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


The solution by tutor @MathLover1 is not correct. Her solution used the intersection of the lines 2x+y=1 and 3x-4y=10 as the point of tangency, which is not right. Her graph shows that her two circles are not quite tangent to the line 3x-4y=10.

Note that the intersection of the two lines would be the point of tangency of the two circles only if the two lines were perpendicular to each other; but they are not.

We need to find point(s) on the line 2x+y=1 that are 5 units from the line 3x-4y=10.

Solve the equation of the line 2x+y=1 for y to get y = -2x+1; then an arbitrary point on that line has coordinates (a,-2a+1).

The distance of a point (p,q) from the line with equation ax+by+c=0 is

abs%28%28ap%2Bbq%2Bc%29%2Fsqrt%28a%5E2%2Bb%5E2%29%29

We want the distance from the point (a,-2a+1) to the line 3x-4y-10=0 to be 5:

abs%28%283%28a%29-4%28-2a%2B1%29-10%29%2F%28sqrt%283%5E2%2B4%5E2%29%29%29=5

%2811a-14%29%2F5=5 OR %2811a-14%29%2F5=-5

11a-14+=+25 OR11a-14+=+-25

11a+=+39 OR 11a+=+-11

a+=+39%2F11 OR +a+=+-1

We have two circles, one on each side of the line 3x-4y=10. The x coordinates of the centers of the circles are a=-1 and a=39/11; the corresponding y coordinates are -2a+1 = 3 and -2a+1 = -78/11+1 = -67/11.

So there are two circles with radius 5 and with centers on the line 2x+y=1 that are tangent to 3x-4y=10.

(1) center (-1,3): %28x%2B1%29%5E2%2B%28y-3%29%5E2+=+25
(2) center (39/11,-67/11): %28x-39%2F11%29%5E2%2B%28y%2B67%2F11%29%5E2+=+25

Here is a graph:



You can see that the two points of tangency are on either side of the line 2x+y=1.