Understanding the Equation of a Circle from Diameter Endpoints

The question asks for the equation of a circle when we are given the coordinates of the endpoints of its diameter. Let these endpoints be $(x_1, y_1)$ and $(x_2, y_2)$.

Geometric Derivation of the Diameter Form

Consider a circle with diameter endpoints $(x_1,y_1)$ and $(x_2,y_2$). Let $(x,y)$ be any point on the circumference of the circle. A key property of a circle is that the angle subtended by the diameter at any point on the circumference is a right angle (90 degrees).

This means the line segment connecting $(x,y)$ to $(x_1,y_1)$ is perpendicular to the line segment connecting $(x,y)$ to $(x_2,y_2)$.

Using Slopes for Perpendicular Lines

The slope of the line joining two points $(x_a,y_a)$ and $(x_b,y_b)$ is given by $\frac{y_b-y_a}{x_b-x_a}$.

The slope of the line segment from $(x_1,y_1)$ to $(x,y)$ is $m_1=\frac{y-y_1}{x-x_1}$.

The slope of the line segment from $(x_2,y_2)$ to $(x,y)$ is $m_2=\frac{y-y_2}{x-x_2}$.

For perpendicular lines (assuming neither is vertical or horizontal), the product of their slopes is -1:

$$m_1\cdot m_2=-1$$

$$\frac{y-y_1}{x-x_1}\cdot \frac{y-y_2}{x-x_2}=-1$$

Multiply both sides by $(x-x_1)(x-x_2)$:

$$(y-y_1)(y-y_2)=-(x-x_1)(x-x_2)$$

Move the term from the right side to the left:

$$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$

This is the required equation of the circle with the given diameter endpoints. This form is sometimes called the diameter form of the circle equation.

Analyzing the Options

Let's compare the derived equation with the given options:

• Option 1: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=x^2+y^2$. This does not match our derived equation.
• Option 2: ${(x-x_1)}^2+{(y-y_1)}^2=x^2{y}^2$. This represents the square of the distance from $(x,y)$ to $(x_1,y_1)$ but set equal to a strange term $x^2{y}^2$. This is not the standard equation of a circle.
• Option 3: $x^2+y^2+2x_1{x}_2+2y_1{y}_2=0$. This does not resemble the standard form or the diameter form of the circle equation. The terms $2x_1{x}_2$ and $2y_1{y}_2$ are constants, making this equation represent a circle centered at the origin with a modified radius, which is not related to the diameter endpoints in this way.
• Option 4: $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$. This exactly matches the equation we derived.

Conclusion

The equation of a circle whose end points of a diameter are $(x_1,y_1)$ and $(x_2,y_2)$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$.

Revision Table: Circle Equation Forms

Additional Information on Circle Properties

• Center: The center of the circle with diameter endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is the midpoint of the diameter. The midpoint coordinates are $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
• Radius: The radius is half the length of the diameter. The length of the diameter can be found using the distance formula between the endpoints $(x_1,y_1)$ and $(x_2,y_2)$, which is $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$. The radius $r$ is $\frac{1}{2}\sqrt{{(x_2-x_1)}^2+{({\mathrm{y<mn>2</mn>}}_{}-y_1)}^2}$.
• Alternative Derivation: You could also find the center and radius using the midpoint and distance formulas, and then plug them into the standard form of the circle equation ${(x-h)}^2+{(y-k)}^2=r^2$. Expanding and simplifying this would lead to the same diameter form.