Question 1141543: Find the center of the circles tangent to 5x-2y-1=0 at 1,2 with a radius of 3
Found 2 solutions by MathLover1, Edwin McCravy: Answer by MathLover1(20849) (Show Source):
You can put this solution on YOUR website! If the equation of the circle is
where ( , ) is the center, we can find the gradient by differentiation:
At ( , ) the gradient is given by
Therefore 
Point ( , ) lies on the circle:
The gradient of the line is given by
: 
This gradient is tangent to the circle, so
So,
=±
= ±
= ( ± )= ± = ±
So we have two circles with different centers:
( , ) and ( , )
or approximately
( , ) and ( , )
equations:
and
Answer by Edwin McCravy(20055) (Show Source):
You can put this solution on YOUR website! Let the center of the circle be (h,k).
The red line is the given line
5x-2y-1 = 0. The green line is the line perpendicular to it.
Since 5x-2y-1 = 0 solved for y is , its slope
is so the green line has slope and since it
goes through (1,2) we use the point slope formula and get the equation:
<-- equation of green line.
Since (h,k) lies on the green line,
Next we use the formula for the distance from a point (x1,y1) to the line
Ax+By+C=0, which is
and the distance from (h,k) must be ±3 units to the line 5x-2y-1=0, so we have
We solve the system:
using the + and then the -, the two centers are
(h,k) = (-1.785439973,3.114172029)
and
(h,k) = (3.785430073,0.8858279709)
Edwin
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