Question

The normal to a given curve at each point (x, y) on the curve passes through the point (3,0). If the curve contains the point (3, 4), then its equation is:

• A. $x^2 + y^2 + 6x - 7 = 0$
• B. $x^2 + y^2 - 6x + 7 = 0$
• C. $x^2 + y^2 - 6x - 7 = 0$
• D. None of these

Answer 1

Answer: (C)

x^2 + y^2 - 6x - 7 = 0

Answer 2

The slope of the normal to a curve at a point (x, y) is given by the slope of the line connecting (x, y) and (3, 0).
Slope of normal ($m_N$) = $\frac{y - 0}{x - 3} = \frac{y}{x - 3}$.

The slope of the tangent to the curve at (x, y) is $\frac{dy}{dx}$.
Since the tangent and normal are perpendicular, the product of their slopes is -1.
$m_T \cdot m_N = -1$
$\frac{dy}{dx} \cdot \left(\frac{y}{x - 3}\right) = -1$

This gives the differential equation:
$\frac{dy}{dx} = -\frac{x - 3}{y}$
Separate the variables:
$y \, dy = -(x - 3) \, dx$
$y \, dy = (3 - x) \, dx$

Integrate both sides:
$\int y \, dy = \int (3 - x) \, dx$
$\frac{y^2}{2} = 3x - \frac{x^2}{2} + C$

Multiply by 2 to clear the denominators:
$y^2 = 6x - x^2 + 2C$
Let $K = 2C$ (a new constant):
$y^2 = 6x - x^2 + K$

Rearrange the terms to form the equation of the curve:
$x^2 + y^2 - 6x = K$

The curve passes through the point (3, 4). Substitute these coordinates into the equation to find the value of K:
$(3)^2 + (4)^2 - 6(3) = K$
$9 + 16 - 18 = K$
$25 - 18 = K$
$K = 7$

Substitute the value of K back into the equation:
$x^2 + y^2 - 6x = 7$
$x^2 + y^2 - 6x - 7 = 0$

This equation represents a circle with center (3, 0) and radius 4.

The final answer is $x^2 + y^2 - 6x - 7 = 0$.