SOLUTION: Find the equation of circle passing through (1;-1) and touching he lines 4x+3y+5=0 and 3x-4y-10=0.

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Question 1109198: Find the equation of circle passing through (1;-1) and touching he lines 4x+3y+5=0 and 3x-4y-10=0.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Interesting problem, and quite challenging....

(Although there might be methods of solution that are much easier...!)

If the circle is tangent to the lines 4x+3y+5=0 and 3x-4y-10=0, then the distance from the center of the circle to each of those lines is the same.
Let the center of the circle be (x,y).

The distance from (x,y) to the line 4x+3y+5=0 is

abs%28%284x%2B3y%2B5%29%2Fsqrt%284%5E2%2B3%5E2%29%29+=+abs%28%284x%2B3y%2B5%29%2F5%29

The distance from (x,y) to the line 3x-4y-10=0 is

abs%28%283x-4y-10%29%2Fsqrt%283%5E2%2B4%5E2%29%29+=+abs%28%283x-4y-10%29%2F5%29

Those distances are the same.

abs%28%284x%2B3y%2B5%29%2F5%29+=+abs%28%283x-4y-10%29%2F5%29
4x%2B3y%2B5+=+3x-4y-10 or 4x%2B3y%2B5+=+-3x%2B4y%2B10
x%2B7y%2B15+=+0 or 7x-y-5+=+0

The center of the circle must lie either on line x+7y+15=0 or on line 7x-y-5=0.

A graph of the two given lines and of the two lines on which the center of the circle must lie is appropriate at this point....



red: 4x+3y+5=0
green: 3x-4y-10=0
blue: x+7y+15=0
purple: 7x-y-5=0

Since the circle must contain the point (1,-1), the graph shows us that the center of the circle must be on the purple line (7x-y-5=0) and not the blue line 9x+7y+15=0).

The graph also shows us that there will be two solutions to the problem -- one with the point (1,-1) near the top of a small circle, and another with the point (1,-1) near the bottom of a larger circle.

Let (x,y) = (x,7x-5) be an arbitrary point on line 7x-y-5=0.

We can first verify that any point on that line is equidistant from the two given lines:

The distance from (x,7x-5) to the line 4x+3y+5=0 is
abs%28%284x%2B3%287x-5%29%2B5%29%2F5%29+=+abs%28%2825x-10%29%2F5%29+=+abs%285x-2%29
The distance from (x,7x-5) to the line 3x-4y-10=0 is
abs%28%283x-4%287x-5%29-10%29%2F5%29+=+abs%28%28-25x%2B10%29%2F5%29+=+abs%28-5x%2B2%29

So any point on the line 7x-y-5=0 is equidistant from the two given lines.
Now we need to find the point(s) on the line where the distance from the point to the given point (1,-1) is the same as the distance to each of the given lines.

Since the distance formula squares the distances, we don't need to concern ourselves with the absolute values.

%28x-1%29%5E2+%2B+%28%287x-5%29-%28-1%29%29%5E2+=+%285x-2%29%5E2
x%5E2-2x%2B1+%2B+49x%5E2-56x%2B16+=+25x%5E2-20x%2B4
25x%5E2-38x%2B13+=+0
%2825x-13%29%28x-1%29+=+0
x+=+13%2F25 or x+=+1

For x = 13/25, y+=+7x-5+=+91%2F25-5+=+-34%2F25; the center of that circle is A(13/25,-34/25).

For x=1, y+=+7x-5+=+2; the center of that circle is B(1,2).

We can verify that each of the points A(13/25,-34/25) and B(1,2) is equidistant from both of the given lines and from the given point (1,-1).

A(13/25,-34/25)....
The distance from line 4x+3y+5=0 is

The distance from line 3x-4y-10=0 is

The distance from (1,-1) is


So the point (13/25,-34/25) is the center of one circle that satisfies the conditions of the problem.

The equation of that circle is

ANSWER #1: %28x-13%2F25%29%5E2%2B%28y%2B34%2F25%29%5E2+=+%283%2F5%29%5E2+=+9%2F25

B(1,2)....
The distance from line 4x+3y+5=0 is

The distance from line 3x-4y-10=0 is
abs%28%283%281%29-4%282%29-10%29%2F5%29+=+abs%28%283-8-10%29%2F5%29+=+abs%28-15%2F5%29+=+3
The distance from (1,-1) is easily seen to be 3.

So the point (1,2) is the center of a second circle that satisfies the conditions of the problem.

The equation of that circle is

ANSWER #2: %28x-1%29%5E2%2B%28y-2%29%5E2+=+3%5E2+=+9

Let's add those two circles to our graph to see that they satisfy the conditions of the problem.