Equation of a Circle: Derivation and Verification

Equation of a circle is a fundamental concept in geometry, often used in various mathematical and engineering applications. This article will explore the derivation of the equation of a circle when the center and a point on the circle are given. Additionally, we will verify the correct equation through substitution and evaluation.

Derivation of the Circle's Equation

Given the center of the circle at (1, -2) and a point on the circle (4, 1), we need to find the equation of the circle.

Step 1: Calculate the Radius

First, find the radius r of the circle using the distance formula between the center and the point:

r sqrt{(4 - 1)^2 (1 - (-2))^2}

Calculate this:

begin{align*} r sqrt{3^2 3^2} sqrt{9 9} sqrt{18} 3sqrt{2} end{align*}

The radius of the circle is 3sqrt{2} or approximately 4.24.

Step 2: Formulate the Circle's Equation

The standard form of the equation of a circle is:

(x - h)^2 (y - k)^2 r^2

Substitute the center coordinates (h, k) (1, -2) and the radius squared r^2 (3sqrt{2})^2 18:

(x - 1)^2 (y 2)^2 18

Simplify and expand this equation:

begin{align*} (x - 1)^2 (y 2)^2 18 (x^2 - 2x 1) (y^2 4y 4) 18 x^2 - 2x 1 y^2 4y 4 18 x^2 y^2 - 2x 4y 5 18 x^2 y^2 - 2x 4y - 13 0 end{align*}

Thus, the equation of the circle is x^2 y^2 - 2x 4y - 13 0.

Verification of the Circle's Equation

To verify, we will substitute the point (4, 1) into the equation and ensure the equality holds:

begin{align*} 4^2 1^2 - 2(4) 4(1) - 13 0 16 1 - 8 4 - 13 0 16 1 4 - 8 - 13 0 21 - 21 0 0 0 end{align*}

Since the point (4, 1) satisfies the equation, the derived equation is correct.

Conclusion

The equation of the circle with a center at (1, -2) and passing through the point (4, 1) is x^2 y^2 - 2x 4y - 13 0. This derivation and verification process ensures the correctness of the equation.

Keyword: circle equation, center and radius, verification