The equation of the circle with radius 3 and centre as the point of intersection of the lines 2x+3y=5,2x-y=1 is

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The equation of the circle with radius 3 and centre as the point of intersection of the lines  2x+3y=5,2x-y=1 is

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Find Center of Circle: To find the center of the circle, we need to solve the system of equations given by the lines 2x+3y=5 and 2x-y=1. Let's solve for x and y using the substitution or elimination method.Eliminate x: First, we can subtract the second equation from the first to eliminate x: (2x + 3y) - (2x - y) = 5 - 1 This simplifies to: 3y + y = 4 4y = 4 Now, we divide by 4 to find the value of y: y = 4 / 4 y = 1Substitute y into Equation: Now that we have the value of y, we can substitute it back into one of the original equations to find x. Let's use the second equation: 2x - y = 1 Substitute y = 1: 2x - 1 = 1 Add 1 to both sides: 2x = 2 Now, divide by 2 to find the value of x: x = 2 / 2 x = 1Calculate x: We have found the center of the circle to be at the point (1, 1). Now, we can write the standard form equation of the circle using the center (h, k) and the radius r. The standard form equation of a circle is: (x - h)^2 + (y - k)^2 = r^2Write Standard Form Equation: Substitute h = 1, k = 1, and r = 3 into the standard form equation: (x - 1)^2 + (y - 1)^2 = 3^2 Simplify the radius squared: (x - 1)^2 + (y - 1)^2 = 9

The equation of the circle with radius $3$ and centre as the point of intersection of the lines $2 \mathrm{x}+3 \mathrm{y}=5,2 \mathrm{x}-\mathrm{y}=1$ is

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The equation of the circle with radius 333 and centre as the point of intersection of the lines 2x+3y=5,2x−y=1 2 \mathrm{x}+3 \mathrm{y}=5,2 \mathrm{x}-\mathrm{y}=1 2x+3y=5,2x−y=1 is

The equation of the circle with radius $3$ and centre as the point of intersection of the lines $2 \mathrm{x}+3 \mathrm{y}=5,2 \mathrm{x}-\mathrm{y}=1$ is