Question 1066857: find the equation of the circle which is tangent to the line 3y-4x-11 and pasese through (8,4) Found 2 solutions by math_helper, MathTherapy:Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website! The problem, as stated, has an infinite number of solutions.
I will find one circle then explain why there are an infinite number of solutions.
The circle I will find is the one tangent to the line 3y-4x-11=0 AND also tangent to a line parallel to this line where this 2nd line passes through (8,4). The circle is tangent to this 2nd line at (8,4). In this way, there is only one solution.
—>
Now imagine a line passing through (8,4), then the center of the circle, then through at a point where the line is tangent to the circle. This (3rd) line has slope -3/4 because it must be perpendicular to (if a line has slope m then a line perpendicular to it has slope -1/m).
The line through the center of the circle also passes through (8,4) so we can find the equation of it.
y = mx+b
4 = (-3/4)(8) + b
4 + (3/4)(8) = b
4 + 6 = b —> b=10
<<< line through center of circle,
perpendicular to
Where these two lines meet, their y values are the same:
=
This reduces to —>
So the diameter of the circle is:
= ——>
The center of the circle is at ( , )
which works out to ( , )
—
The equation of the circle is therefore:
—
Now, the original problem statement has an infinite number of solutions because you can make a circle tangent to that is a little bigger than the one we just found above, and passes through (8,4) but at a different point on the circle than the line perpendicular to . This can be done an infinite number of times by moving the tangent point (& adjusting circle size) by an infinitesimally small amount.
—