Introduction to the Concept of an Equation of a Circle
" "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.
" "Step-by-Step Solution: Finding the Center of the Circle
" "To begin, we need to find the center of the circle, which is the point of intersection of two given lines.
" "Given System of Linear Equations
" "The given system of linear equations is:
" "x - y 4 (1)
" "2x 3y -7 (2)
" "Solving for Intersection Point (Center of the Circle)
" "Step 1: Express the first equation in terms of y.
" "x - y 4 rarr; y x - 4 (3)
" "Step 2: Substitute y from equation (3) into equation (2).
" "2x 3(x - 4) -7
" "2x 3x - 12 -7
" "5x 5
x 1
Step 3: Substitute x 1 back into equation (3).
" "y 1 - 4 -3
" "Thus, the center of the circle is the point (1, -3).
" "Determining the Radius of the Circle
" "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).
" "The formula for the distance between two points (x1, y1) and (x2, y2) is given by:
" "Distance √((x2 - x1)2 (y2 - y1)2)
" "In this case, (x1, y1) (1, -3) and (x2, y2) (4, 8).
" "Step 1: Calculate the difference in x and y coordinates.
" "dx 4 - 1 3
" "dy 8 - (-3) 8 3 11
" "Step 2: Calculate the square of these differences.
" "dx2 32 9
" "dy2 112 121
" "Step 3: Sum the squares of the differences.
" "dx2 dy2 9 121 130
" "Step 4: Take the square root to find the radius.
" "Radius √130
" "Deriving the Equation of the Circle
" "The general equation of a circle with center (h, k) and radius r is given by:
" "(x - h)2 (y - k)2 r2
" "Substitute (h, k) (1, -3) and r √130 into the equation.
" "(x - 1)2 (y 3)2 130
" "Alternatively, you can expand and simplify the equation:
" "x2 - 2x 1 y2 6y 9 130
x2 y2 - 2x 6y 10 130
x2 y2 - 2x 6y - 120 0
Conclusion
" "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:
" "(x - 1)2 (y 3)2 130
" "or
" "x2 y2 - 2x 6y - 120 0
" "Understanding these steps and the underlying math can greatly enhance your problem-solving skills in analytical geometry and calculus.
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