Coordinate Geometry with Tangency

Question

Find the Equation of the circle which is tangent to the x and y-axis and the line 3x+4y=12 at the point (1.6,1.8)

Answer 1

From 3x+4y=12

We need to get the perpendicular line that passes through (1.6,1.8)

4y = -3x + 12

y = (-3/4)x + 3

That slope of 3x+4y=12 is -3/4, therefore the slope of the line perpendicular to it is 4/3. Therefore the equation of the line that we are looking for in point slope form is:

y-1.8 = (4/3)(x-1.6)

and the center of the circle lies there.

The conditions that the circle in first quadrant is both tangent to the x-axis and y-axis are:

1. the coordinates of points of tangency are (x,0) and (0,y)
2. y=x
3. The coordinates of the center is (x,y).
4. The length of the radius r = x = y

So for the equation y-1.8 = (4/3)(x-1.6) and since y=x, let's use y instead of x:

y-1.8 = (4/3)(y-1.6)

y-1.8 = (4/3)y - 32/15

y-(4/3)y = 1.8-32/15

-(1/3)y = -(1/3)

y=1

Since y=x, then x=1.

Therefore the coordinates of the center is (1,1) and the radius is 1.

So for the equation of the circle in standard form (x-h)² + (y-k)² = r²

(x-1)² + (y-1)² = 1

Let's check also if the distance from (1.6,1.8) to (1,1) is also 1:

r²= (1-1.6)² + (1-1.8)²

r²= (-.6)² + (-.8)² = .36 + .64

r²= 1.00

r = 1

Therefore our answer is correct.

Answer 2

The given line has a slope of m = - 3/4, which means the radius thru (1.6, 1.8) will have a slope of m = 4/3 (opposite reciprocal for a perpendicular line, which is true of the radius and a tangent at the pt of tangency).

So the center of the circle lies on the line with equation y - 1.8 = 4/3 (x - 1.6).

For the circle to be tangent to both axes, the center must also lie on the line y = x: x - 1.8 = 4/3 (x - 1.6)

x - 9/5 = 4/3x - 32/15

1/3x = 1/3

x = 1 , y = 1 , Ctr: (1 , 1) , r = 1

(x - 1)² + (y - 1)² = 1. (We leave the proof that (1.6, 1.8) lies on the circle and is the pt of tangency between the circle and given line as an exercise for the reader.)