Understanding Tangency in Geometry
In coordinate geometry, a line is considered tangent to a circle if it touches the circle at exactly one point. A fundamental property of a tangent line is that the perpendicular distance from the center of the circle to the line is exactly equal to the radius of the circle.
The Mathematical Problem
Given:
- The circle equation: $x^2 + y^2 = a^2$
- The line equation: $px + qy = r$
We need to find the condition that makes the line tangent to the circle.
Step-by-Step Derivation
1. Identify the properties of the circle
For the circle $x^2 + y^2 = a^2$:
- The center is at the origin $(0, 0)$.
- The radius is $a$.
2. Use the perpendicular distance formula
The perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
3. Rewrite the line equation
To apply the formula, we express the line $px + qy = r$ as: $px + qy - r = 0$ Here, $A = p$, $B = q$, and $C = -r$.
4. Set the distance equal to the radius
Since the line is a tangent, the distance from the center $(0,0)$ to the line must equal the radius $a$: $a = \frac{|p(0) + q(0) - r|}{\sqrt{p^2 + q^2}}$
5. Simplify the expression
$a = \frac{|-r|}{\sqrt{p^2 + q^2}}$ $a = \frac{|r|}{\sqrt{p^2 + q^2}}$
Squaring both sides to remove the absolute value and the square root: $a^2 = \frac{r^2}{p^2 + q^2}$
By cross-multiplying, we get the required condition: $r^2 = a^2(p^2 + q^2)$
Conclusion
The condition for the line $px + qy = r$ to be tangent to the circle $x^2 + y^2 = a^2$ is $r^2 = a^2(p^2 + q^2)$. This relationship ensures that the line is situated at a distance of exactly 'a' units from the origin, confirming it touches the boundary of the circle perfectly.