Answer :
To solve the given system of equations:
[tex]\[ \begin{cases} x^2 + y^2 = 4 \\ x - y = 1 \end{cases} \][/tex]
we have to find the coordinates [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously. Let's proceed step-by-step:
1. Equation of the Circle:
[tex]\[ x^2 + y^2 = 4 \][/tex]
This represents a circle with a radius of [tex]\(2\)[/tex] centered at the origin [tex]\((0,0)\)[/tex].
2. Equation of the Line:
[tex]\[ x - y = 1 \][/tex]
This is a straight line with a slope of [tex]\(1\)[/tex] and y-intercept of [tex]\(-1\)[/tex].
To find the intersection points of the circle and the line, we need to solve these equations together.
3. Substitution Method:
From the second equation [tex]\(x - y = 1\)[/tex], we can solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = y + 1 \][/tex]
4. Substitute [tex]\(x = y + 1\)[/tex] into the first equation [tex]\(x^2 + y^2 = 4\)[/tex]:
[tex]\[ (y + 1)^2 + y^2 = 4 \][/tex]
5. Expand and simplify:
[tex]\[ y^2 + 2y + 1 + y^2 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 - 4 = 0 \][/tex]
[tex]\[ 2y^2 + 2y - 3 = 0 \][/tex]
6. Divide through by 2 to simplify:
[tex]\[ y^2 + y - \frac{3}{2} = 0 \][/tex]
7. Solve this quadratic equation for [tex]\(y\)[/tex] using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -\frac{3}{2}\)[/tex].
[tex]\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -\frac{3}{2}}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{1 + 6}}{2} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{7}}{2} \][/tex]
8. Therefore, we get two solutions for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-1 + \sqrt{7}}{2} \quad \text{and} \quad y = \frac{-1 - \sqrt{7}}{2} \][/tex]
9. Now, substitute these [tex]\(y\)[/tex] values back into [tex]\(x = y + 1\)[/tex] to get the corresponding [tex]\(x\)[/tex] values:
[tex]\[ x = \frac{-1 + \sqrt{7}}{2} + 1 = \frac{-1 + \sqrt{7}}{2} + \frac{2}{2} = \frac{1 + \sqrt{7}}{2} \][/tex]
[tex]\[ x = \frac{-1 - \sqrt{7}}{2} + 1 = \frac{-1 - \sqrt{7}}{2} + \frac{2}{2} = \frac{1 - \sqrt{7}}{2} \][/tex]
10. Thus, the solutions for the system of equations are:
[tex]\[ (x, y) = \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right) \quad \text{and} \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right) \][/tex]
These represent the points of intersection between the circle [tex]\(x^2 + y^2 = 4\)[/tex] and the line [tex]\(x - y = 1\)[/tex].
[tex]\[ \begin{cases} x^2 + y^2 = 4 \\ x - y = 1 \end{cases} \][/tex]
we have to find the coordinates [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously. Let's proceed step-by-step:
1. Equation of the Circle:
[tex]\[ x^2 + y^2 = 4 \][/tex]
This represents a circle with a radius of [tex]\(2\)[/tex] centered at the origin [tex]\((0,0)\)[/tex].
2. Equation of the Line:
[tex]\[ x - y = 1 \][/tex]
This is a straight line with a slope of [tex]\(1\)[/tex] and y-intercept of [tex]\(-1\)[/tex].
To find the intersection points of the circle and the line, we need to solve these equations together.
3. Substitution Method:
From the second equation [tex]\(x - y = 1\)[/tex], we can solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = y + 1 \][/tex]
4. Substitute [tex]\(x = y + 1\)[/tex] into the first equation [tex]\(x^2 + y^2 = 4\)[/tex]:
[tex]\[ (y + 1)^2 + y^2 = 4 \][/tex]
5. Expand and simplify:
[tex]\[ y^2 + 2y + 1 + y^2 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 - 4 = 0 \][/tex]
[tex]\[ 2y^2 + 2y - 3 = 0 \][/tex]
6. Divide through by 2 to simplify:
[tex]\[ y^2 + y - \frac{3}{2} = 0 \][/tex]
7. Solve this quadratic equation for [tex]\(y\)[/tex] using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
Here, [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -\frac{3}{2}\)[/tex].
[tex]\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot -\frac{3}{2}}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{1 + 6}}{2} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{7}}{2} \][/tex]
8. Therefore, we get two solutions for [tex]\(y\)[/tex]:
[tex]\[ y = \frac{-1 + \sqrt{7}}{2} \quad \text{and} \quad y = \frac{-1 - \sqrt{7}}{2} \][/tex]
9. Now, substitute these [tex]\(y\)[/tex] values back into [tex]\(x = y + 1\)[/tex] to get the corresponding [tex]\(x\)[/tex] values:
[tex]\[ x = \frac{-1 + \sqrt{7}}{2} + 1 = \frac{-1 + \sqrt{7}}{2} + \frac{2}{2} = \frac{1 + \sqrt{7}}{2} \][/tex]
[tex]\[ x = \frac{-1 - \sqrt{7}}{2} + 1 = \frac{-1 - \sqrt{7}}{2} + \frac{2}{2} = \frac{1 - \sqrt{7}}{2} \][/tex]
10. Thus, the solutions for the system of equations are:
[tex]\[ (x, y) = \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right) \quad \text{and} \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right) \][/tex]
These represent the points of intersection between the circle [tex]\(x^2 + y^2 = 4\)[/tex] and the line [tex]\(x - y = 1\)[/tex].
