Prove that two circles are congruent if their radii are equal

Question

Is this to be proved by showing that the circumferences/areas are equal?

Answer 1

Equation of a circle is x^2+y^2=k. where only parameter on which a circle can be different from another is it's radius K^1/2. Since radius is same every property including area and circumference will be same. This proves the result.

Answer 2

A circle has two properties

1. the location of its center, e.g. $(x_0, y_0)$ and
2. its radius, e.g. $r$

It can be described algebraically as set

$$ C = \left\{ (x, y) \in \mathbb{R} \, \left| \, (x-x_0)^2+(y-y_0)^2=r^2 \right. \right\} \quad (*) $$

For the matter of congruence the location is not relevant, think of arranging two circles such that they have the same center location, this leave the radius as characterizing property, which must be equal to result in the same circle points:

Assume a circle $C_1$ with origin $(x_0, y_0)$ and radius $r_1$, and a circle $C_2$ with same origin $(x_0, y_0)$ and radius $r_2$. Let $(x^*, y^*)$ be an arbitrary point from $C_1$. Then it fulfills the equation

$$ (x^*-x_0)^2+(y^*-y_0)^2=r_1^2 $$

In the case of congruency, it must also be a point of $C_2$, and thus fulfills

$$ (x^*-x_0)^2+(y^*-y_0)^2=r_2^2 $$

Equating both left hand sides gives

$$ r_1^2 = r_2^2 $$

which implies $r_1 = r_2$, for non-negative radii.

Derivation of the Circle Equation

If $u_0 = (x_0, y_0)$ is a vector to the center of a circle, then each point on the circle is given by the the vector $u = (x, y)$, while the arrow from the center of the circle to the point on the circle is given by the difference vector $u - u_0$.

All points on the circle stay the same distance from the center away, thus fulfill:

$$ \begin{align} r &= |u - u_0| \\ &= \sqrt{(u-u_0) \cdot (u - u_0)} \\ &= \sqrt{(u - u_0)_x^2 + (u - u_0)_y^2} \\ &= \sqrt{(x - x_0)^2 + (y - y_0)^2} \end{align} $$

where the scalar product is used as definition of lengths and angles. Squaring this equation on both sides gives equation (*).

If Linear Algebra is not used, it boils down to using the numbers

$$ (x - x_0, y - y_0) $$

for the length of the catheti of the triangle with right angle and applying the theorem of Pythagoras to state $$ r^2 = (x -x_0)^2 + (y - y_0)^2 $$