Question

Find the differential equation of all circles passing through the origin and having their centres on the x-axis.

Answer 1

Step 1: General Equation of a Circle:
The general equation of a circle with center (h, k) and radius r is:
(x - h)² + (y - k)² = r²

Step 2: Apply the Given Conditions:
* Circle passes through the origin (0, 0):
Substituting (0, 0) into the general equation:
(0 - h)² + (0 - k)² = r²
h² + k² = r²

* Center lies on the x-axis:
This means the y-coordinate of the center is zero, so k = 0.

Step 3: Substitute k = 0 into the Equation:
Substituting k = 0 into h² + k² = r²:
h² + 0² = r²
h² = r²

Step 4: Simplify the General Equation:
Now, substitute k = 0 and r² = h² into the general equation of the circle:

(x - h)² + (y - 0)² = h²
(x - h)² + y² = h²
x² - 2xh + h² + y² = h²
x² - 2xh + y² = 0

Step 5: Differentiate with Respect to x:
Differentiate the equation x² - 2xh + y² = 0 with respect to x:
2x - 2(h * 1 + x * dh/dx) + 2y * dy/dx = 0
2x - 2h - 2x * dh/dx + 2y * dy/dx = 0

Since h is a parameter, we can consider it a constant. However, we can also see that it is a function of x and y.

But from the original equation: 2xh = x² + y², so h=(x²+y²)/(2x)

Substituting h back into the differentiated equation.
2x - 2((x² + y²)/(2x)) - 2x * d/dx((x² + y²)/(2x)) + 2y * dy/dx = 0;
2x - (x² + y²)/x - 2x * ((2x * 2x - (x²+y²) * 2)/(4x²)) + 2y * dy/dx = 0
2x - (x² + y²)/x - (4x² - 2x²-2y²)/(2x) + 2y*dy/dx = 0
2x - (x² + y²)/x - (2x²-2y²)/(2x) + 2y*dy/dx = 0
2x - (x² + y²)/x - (x²-y²)/x + 2y*dy/dx = 0
2x - (2x²)/x + 2y*dy/dx = 0
2x - 2x + 2y*dy/dx = 0
2y*dy/dx=0
y*dy/dx=0

This is incorrect, let us use the other method.

From x² - 2xh + y² = 0.
2xh = x² + y²
h = (x² + y²)/(2x)

Differentiate with respect to x.
2xh = x² + y²

2h + 2x(dh/dx) = 2x + 2y(dy/dx)
dh/dx = (2x + 2y(dy/dx) - 2h)/(2x)
dh/dx = (x + y(dy/dx) - h)/x

Substituting h = (x²+y²)/(2x)
dh/dx = (x+y(dy/dx)-(x²+y²)/(2x))/x
dh/dx = (2x² + 2xy(dy/dx)-x²-y²)/(2x²)
dh/dx = (x² + 2xy(dy/dx)-y²)/(2x²)

Now, 2x - 2h - 2x(dh/dx) + 2y(dy/dx)=0
2x-2(x²+y²)/(2x)-2x(x² + 2xy(dy/dx)-y²)/(2x²) + 2y(dy/dx) = 0
2x-(x²+y²)/x - (x²+2xy(dy/dx)-y²)/x + 2y(dy/dx) = 0
2x² - (x²+y²) - (x²+2xy(dy/dx)-y²) + 2xy(dy/dx) = 0
2x² - x² - y² - x² - 2xy(dy/dx) + y² + 2xy(dy/dx) = 0
0=0

This method is also not working.

We will use implicit differentiation.
x² -2xh + y² = 0
2x-2h-2x(dh/dx) + 2y(dy/dx) = 0
2x - 2h = 2(x(dh/dx) - y(dy/dx))

h = (x² + y²)/2x
dh/dx = (2x * 2x - 2(x²+y²))/(4x²) = (4x² - 2x²-2y²)/(4x²) = (2x²-2y²)/(4x²) = (x²-y²)/(2x²)

2x - 2(x²+y²)/2x = 2(x(x²-y²)/(2x²) - y(dy/dx))
2x - (x²+y²)/x = (x²-y²)/x - 2y(dy/dx)
2x² - x² - y² = x² - y² - 2xy(dy/dx)
0 = -2xy(dy/dx)
x(dy/dx) + y = 0

Final Answer:
x(dy/dx) + y = 0