document.write( "Question 1097107: The line 4x+3y=25 & 3x-4y=-7 are tangent to a circle. Connecting the center of the circle, point of tangencies, and intersection of the 2 lines will form a square. The center of the circle is 6 units from the intersection of the 2 lines. Find the equation of the circle. \n" ); document.write( "
Algebra.Com's Answer #711707 by htmentor(1343)\"\" \"About 
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First, we need to find the intersection point of the two lines
\n" ); document.write( "4x+3y=25 -> 16/3x + 4y = 100/3 [multiply by 4/3]
\n" ); document.write( "3x-4y=-7
\n" ); document.write( "25/3x = 100/3 - 21/3 = 79/3 = 3.16
\n" ); document.write( "x = 79/25
\n" ); document.write( "y = (25 - 4(79/25))/3 = 309/75 = 4.12
\n" ); document.write( "The intersection point is A = (3.16,4.12)
\n" ); document.write( "The length of the line segment AC, where C is the center point of the circle, = 6 units
\n" ); document.write( "The segment AC is a diagonal of the square connecting A, C and the two tangent points B and D
\n" ); document.write( "The angle between AC and CB and CD is therefore equal to 45 degrees,
\n" ); document.write( "since AC is a bisector of the two tangent lines meeting at A
\n" ); document.write( "Hence the segments CB and CD are 6/sqrt(2) in length, and this is equal to the radius of the circle.
\n" ); document.write( "Now we need to find the center of the circle.
\n" ); document.write( "Lines which are parallel to the tangent lines, and are 6/sqrt(2) units away, will go through the center of the circle.
\n" ); document.write( "The perpendicular distance between two parallel lines can be written as
\n" ); document.write( "d = |b2-b1|/sqrt(m^2+1), where m is the slope and b1, b2 the intercepts
\n" ); document.write( "In standard form 4x+3y=25 is y = -4/3x + 25/3, m = -4/3 b1 = 25/3
\n" ); document.write( "We can use this to solve for b2
\n" ); document.write( "d^2 = 18 = (25/3-b2)^2/((-4/3)^2+1)) = ((25/3)^2 - 50/3b + b2^2)/(25/9)
\n" ); document.write( "9/25b2^2 - 6b2 + 7 = 0
\n" ); document.write( "Solving for b2 gives 1.2623 and 15.4044, which means there are two circles which meet the conditions.
\n" ); document.write( "Take b2 = 1.2623. Thus one equation of a line passing through the center is
\n" ); document.write( "y = (-4/3)x + 1.2623 [1]
\n" ); document.write( "Similarly, we can find the equation of the line parallel to 3x-4y=-7, and 6/sqrt(2) units away.
\n" ); document.write( "These two lines will intersect at the center of the circle.
\n" ); document.write( "The equation of this line is y = 3/4x - 3.5533 [2]
\n" ); document.write( "Find the intersection point of [1] and [2]:
\n" ); document.write( "3/4x - 3.5533 = -4/3x + 1.2623 -> x = 2.3115
\n" ); document.write( "Thus y = 3/4*2.3115 - 3.5533 = -1.8197
\n" ); document.write( "Finally, the equation (of one) of the circle(s) is
\n" ); document.write( "(x-2.3115)^2 + (y+1.8197)^2 = 18\r
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