Finding the Equation of a Circle through a Given Point

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Introduction to the Concept of an Equation of a Circle

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An equation of a circle is a mathematical representation that captures the locus of points moving in such a manner that the distance to a fixed point, termed the center, is always a constant. This article explores the process of determining the equation of a circle that passes through a specific point, using the intersection of two lines as the center of the circle. We will provide detailed steps, mathematical explanations, and practical examples to ensure a comprehensive understanding.

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Step-by-Step Solution: Finding the Center of the Circle

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To begin, we need to find the center of the circle, which is the point of intersection of two given lines.

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Given System of Linear Equations

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The given system of linear equations is:

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x - y 4 (1)

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2x 3y -7 (2)

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Solving for Intersection Point (Center of the Circle)

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Step 1: Express the first equation in terms of y.

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x - y 4 rarr; y x - 4 (3)

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Step 2: Substitute y from equation (3) into equation (2).

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2x 3(x - 4) -7

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2x 3x - 12 -7

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5x 5
x 1

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Step 3: Substitute x 1 back into equation (3).

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y 1 - 4 -3

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Thus, the center of the circle is the point (1, -3).

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Determining the Radius of the Circle

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To find the equation of the circle, we need to determine the radius. The radius is the distance from the center of the circle to the given point (4, 8).

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The formula for the distance between two points (x1, y1) and (x2, y2) is given by:

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Distance √((x2 - x1)2 (y2 - y1)2)

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In this case, (x1, y1) (1, -3) and (x2, y2) (4, 8).

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Step 1: Calculate the difference in x and y coordinates.

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dx 4 - 1 3

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dy 8 - (-3) 8 3 11

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Step 2: Calculate the square of these differences.

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dx2 32 9

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dy2 112 121

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Step 3: Sum the squares of the differences.

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dx2 dy2 9 121 130

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Step 4: Take the square root to find the radius.

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Radius √130

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Deriving the Equation of the Circle

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The general equation of a circle with center (h, k) and radius r is given by:

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(x - h)2 (y - k)2 r2

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Substitute (h, k) (1, -3) and r √130 into the equation.

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(x - 1)2 (y 3)2 130

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Alternatively, you can expand and simplify the equation:

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x2 - 2x 1 y2 6y 9 130
x2 y2 - 2x 6y 10 130
x2 y2 - 2x 6y - 120 0

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Conclusion

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We have successfully derived the equation of the circle that passes through the point (4, 8) with its center at the intersection of the lines given by the equations x - y 4 and 2x 3y -7. The final equations are:

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(x - 1)2 (y 3)2 130

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or

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x2 y2 - 2x 6y - 120 0

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Understanding these steps and the underlying math can greatly enhance your problem-solving skills in analytical geometry and calculus.

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