Understanding Tangents and Normals
In coordinate geometry, the tangent to a circle at a specific point is a straight line that touches the circle at exactly that point. The normal is a line perpendicular to the tangent, passing through the same point of tangency. Because the normal is perpendicular to the tangent at the point of contact, it must also pass through the center of the circle.
The Problem
We are given the circle equation: $x^2 + y^2 - 2x - 4y + 3 = 0$ We need to find the equations of the tangent and normal at the point $(2, 3)$.
Step-by-Step Solution
1. Differentiate to find the slope of the tangent
The slope of the tangent ($m_t$) is the derivative of the circle equation with respect to $x$, evaluated at $(2, 3)$.
Differentiating $x^2 + y^2 - 2x - 4y + 3 = 0$ implicitly: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(4y) + \frac{d}{dx}(3) = 0$ $2x + 2y \frac{dy}{dx} - 2 - 4 \frac{dy}{dx} = 0$
Group the $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(2y - 4) = 2 - 2x$ $\frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1 - x}{y - 2}$
2. Calculate the slope at (2, 3)
Substitute $x=2$ and $y=3$ into the derivative: $m_t = \frac{1 - 2}{3 - 2} = \frac{-1}{1} = -1$
3. Equation of the Tangent
Using the point-slope form $(y - y_1) = m(x - x_1)$: $y - 3 = -1(x - 2)$ $y - 3 = -x + 2$ $x + y - 5 = 0$
4. Equation of the Normal
The normal is perpendicular to the tangent, so its slope $m_n$ is the negative reciprocal of $m_t$: $m_n = -\frac{1}{m_t} = -\frac{1}{-1} = 1$
Using the point-slope form again: $y - 3 = 1(x - 2)$ $y - 3 = x - 2$ $x - y + 1 = 0$
Summary
- The equation of the tangent is $x + y - 5 = 0$.
- The equation of the normal is $x - y + 1 = 0$.