Question 1109198
<br>
Interesting problem, and quite challenging....<br>
(Although there might be methods of solution that are much easier...!)<br>
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).<br>
The distance from (x,y) to the line 4x+3y+5=0 is<br>
{{{abs((4x+3y+5)/sqrt(4^2+3^2)) = abs((4x+3y+5)/5)}}}<br>
The distance from (x,y) to the line 3x-4y-10=0 is<br>
{{{abs((3x-4y-10)/sqrt(3^2+4^2)) = abs((3x-4y-10)/5)}}}<br>
Those distances are the same.<br>
{{{abs((4x+3y+5)/5) = abs((3x-4y-10)/5)}}}
{{{4x+3y+5 = 3x-4y-10}}} or {{{4x+3y+5 = -3x+4y+10}}}
{{{x+7y+15 = 0}}} or {{{7x-y-5 = 0}}}<br>
The center of the circle must lie either on line x+7y+15=0 or on line 7x-y-5=0.<br>
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....<br>
{{{graph(400,400,-4,4,-4,4,(-4x-5)/3,(3x-10)/4,(-x-15)/7,7x-5)}}}<br>
red: 4x+3y+5=0
green: 3x-4y-10=0
blue: x+7y+15=0
purple: 7x-y-5=0<br>
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).<br>
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.<br>
Let (x,y) = (x,7x-5) be an arbitrary point on line 7x-y-5=0.<br>
We can first verify that any point on that line is equidistant from the two given lines:<br>
The distance from (x,7x-5) to the line 4x+3y+5=0 is
{{{abs((4x+3(7x-5)+5)/5) = abs((25x-10)/5) = abs(5x-2)}}}
The distance from (x,7x-5) to the line 3x-4y-10=0 is
{{{abs((3x-4(7x-5)-10)/5) = abs((-25x+10)/5) = abs(-5x+2)}}}<br>
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.<br>
Since the distance formula squares the distances, we don't need to concern ourselves with the absolute values.<br>
{{{(x-1)^2 + ((7x-5)-(-1))^2 = (5x-2)^2}}}
{{{x^2-2x+1 + 49x^2-56x+16 = 25x^2-20x+4}}}
{{{25x^2-38x+13 = 0}}}
{{{(25x-13)(x-1) = 0}}}
{{{x = 13/25}}} or {{{x = 1}}}<br>
For x = 13/25, {{{y = 7x-5 = 91/25-5 = -34/25}}}; the center of that circle is A(13/25,-34/25).<br>
For x=1, {{{y = 7x-5 = 2}}}; the center of that circle is B(1,2).<br>
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).<br>
A(13/25,-34/25)....
The distance from line 4x+3y+5=0 is
{{{abs((4(13/25)+3(-34/25)+5)/5) = abs((52/25-102/25+5)/5) = abs((-50/25+5)/5) = abs((-2+5)/5) = 3/5}}}
The distance from line 3x-4y-10=0 is
{{{abs((3(13/25)-4(-34/25)-10)/5) = abs((39/25+136/25-10)/5) = abs((175/25-10)/5) = abs((7-10)5) = 3/5}}}
The distance from (1,-1) is
{{{sqrt((13/25-1)^2+(-34/25+1)^2) = sqrt((12/25)^2+(-9/25)^2) = 15/25 = 3/5}}}<br>
So the point (13/25,-34/25) is the center of one circle that satisfies the conditions of the problem.<br>
The equation of that circle is<br>
ANSWER #1: {{{(x-13/25)^2+(y+34/25)^2 = (3/5)^2 = 9/25}}}<br>
B(1,2)....
The distance from line 4x+3y+5=0 is
{{{abs((4(1)+3(2)+5)/5) = abs((4+6+5)/5) = abs(15/5) = 3}}}
The distance from line 3x-4y-10=0 is
{{{abs((3(1)-4(2)-10)/5) = abs((3-8-10)/5) = abs(-15/5) = 3}}}
The distance from (1,-1) is easily seen to be 3.<br>
So the point (1,2) is the center of a second circle that satisfies the conditions of the problem.<br>
The equation of that circle is<br>
ANSWER #2: {{{(x-1)^2+(y-2)^2 = 3^2 = 9}}}<br>
Let's add those two circles to our graph to see that they satisfy the conditions of the problem.<br>
{{{graph(400,400,-4,4,-4,4,(-4x-5)/3,(3x-10)/4,7x-5,sqrt((9/25-(x-13/25)^2))-34/25,-sqrt((9/25-(x-13/25)^2))-34/25,sqrt(9-(x-1)^2)+2,-sqrt(9-(x-1)^2)+2)}}}<br>