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

Full solution

Q. 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

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: $\newline$ $(2x + 3y) - (2x - y) = 5 - 1$ $\newline$ This simplifies to: $\newline$ $3y + y = 4$ $\newline$ $4y = 4$ $\newline$ Now, we divide by $4$ to find the value of y: $\newline$ $y = \frac{4}{4}$ $\newline$ $y = 1$
Substitute $y$ 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: $\newline$ $2x - y = 1$ $\newline$ Substitute $y = 1$ : $\newline$ $2x - 1 = 1$ $\newline$ Add $1$ to both sides: $\newline$ $2x = 2$ $\newline$ Now, divide by $2$ to find the value of $x$ : $\newline$ $y$ $0$ $\newline$ $y$ $1$
Calculate $x$ : 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^ $2$
Write Standard Form Equation: Substitute $h = 1$ , $k = 1$ , and $r = 3$ into the standard form equation: $\newline$ $(x - 1)^2 + (y - 1)^2 = 3^2$ $\newline$ Simplify the radius squared: $\newline$ $(x - 1)^2 + (y - 1)^2 = 9$