Question

Find the differential equation of all circles in the first quadrant which touch both the coordinate axes.

Hint:

Use implicit differentiation to find the slope of tangents and eliminate parameters by substitution.

Answer 1

Step 1: The general equation of a circle touching both axes is: \[ (x - r)^2 + (y - r)^2 = r^2, \] where \( r \) is the radius of the circle, and the center is at \( (r, r) \). 
Step 2: Differentiate implicitly with respect to \( x \): \[ 2(x - r) + 2(y - r) \frac{dy}{dx} = 0 \quad \Rightarrow \quad (x - r) + (y - r) \frac{dy}{dx} = 0. \] 
Step 3: Eliminate \( r \) using the relationship \( x^2 + y^2 = 2xr \): Substitute \( r = \frac{x^2 + y^2}{2x} \) into the equation: \[ \frac{dy}{dx} = -\frac{x - \frac{x^2 + y^2}{2x}}{y - \frac{x^2 + y^2}{2x}}. \]