Understanding Tangents and Normals to a Circle
In coordinate geometry, a tangent line to a circle at a specific point is a line that touches the circle at exactly that point. A normal line is perpendicular to the tangent line at the point of contact and always passes through the center of the circle.
The Problem
Find the equations of the tangent and normal to the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the point $(2, 3)$.
Step-by-Step Solution
1. Differentiate the Circle's Equation
To find the slope of the tangent at point $(x, y)$, we use implicit differentiation with respect to $x$:
$\frac{d}{dx}(x^2 + y^2 - 2x - 4y + 3) = \frac{d}{dx}(0)$ $2x + 2y\frac{dy}{dx} - 2 - 4\frac{dy}{dx} = 0$
Group the $\frac{dy}{dx}$ terms: $(2y - 4)\frac{dy}{dx} = 2 - 2x$ $\frac{dy}{dx} = \frac{2 - 2x}{2y - 4} = \frac{1 - x}{y - 2}$
2. Calculate the Slope of the Tangent ($m$)
At the point $(2, 3)$, we substitute $x = 2$ and $y = 3$ into the derivative: $m = \frac{1 - 2}{3 - 2} = \frac{-1}{1} = -1$
3. Equation of the Tangent Line
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 Line
The normal is perpendicular to the tangent. If the tangent slope is $m = -1$, the normal slope $m'$ is $-\frac{1}{m} = 1$.
Using point-slope form again: $y - 3 = 1(x - 2)$ $y - 3 = x - 2$ $x - y + 1 = 0$
Summary
- Tangent Equation: $x + y - 5 = 0$
- Normal Equation: $x - y + 1 = 0$
Intuition
- The tangent represents the instantaneous direction of the curve at that point.
- The normal is the line segment connecting the center of the circle $(1, 2)$ to the point on the circumference $(2, 3)$. You can verify that the normal line $x - y + 1 = 0$ passes through the center $(1, 2)$ by plugging in the values: $1 - 2 + 1 = 0$.