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) (Show Source): You can put this solution on YOUR website!
find intersection of line and tangent
.......solve for
---------------------
substitute in
then,
intersection point is:
P( ,)
-> since center is on a line , we have
=>
so, center C (,)=(,)
distance from and is
or
=>
or
so, we have two centers that lie on given line:
center C (, )
and
center C (,)
then, we have two circles:
or
Answer by greenestamps(13203) (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
We want the distance from the point (a,-2a+1) to the line 3x-4y-10=0 to be 5:
OR
OR
OR
OR
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):
(2) center (39/11,-67/11):
Here is a graph:
You can see that the two points of tangency are on either side of the line 2x+y=1.
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