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Sagot :
To find the coordinates of point B, we need to solve the system of equations given by the circle [tex]\( C \)[/tex] and the line [tex]\( L \)[/tex].
1. Equation of the Circle:
The circle is centered at the origin and has the equation:
[tex]\[ x^2 + y^2 = 20 \][/tex]
2. Equation of the Line:
The line has the equation:
[tex]\[ x + 2y = 5 \][/tex]
3. Solving the System of Equations:
We solve these two equations simultaneously to find the points of intersection (where the line intersects the circle).
4. Substitution Method:
From the line equation [tex]\( x + 2y = 5 \)[/tex], solve for [tex]\( x \)[/tex]:
[tex]\[ x = 5 - 2y \][/tex]
5. Substitute into the Circle Equation:
Substitute [tex]\( x = 5 - 2y \)[/tex] into the circle equation [tex]\( x^2 + y^2 = 20 \)[/tex]:
[tex]\[ (5 - 2y)^2 + y^2 = 20 \][/tex]
6. Expand and Simplify:
[tex]\[ (5 - 2y)^2 = 25 - 20y + 4y^2 \][/tex]
So, the equation becomes:
[tex]\[ 25 - 20y + 4y^2 + y^2 = 20 \][/tex]
Combine like terms:
[tex]\[ 5y^2 - 20y + 25 = 20 \][/tex]
Subtract 20 from both sides:
[tex]\[ 5y^2 - 20y + 5 = 0 \][/tex]
7. Simplify the Quadratic Equation:
[tex]\[ y^2 - 4y + 1 = 0 \][/tex]
8. Solve the Quadratic Equation (Use the Quadratic Formula):
Recall, the quadratic formula is [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
For our quadratic equation [tex]\( y^2 - 4y + 1 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ y = 2 \pm \sqrt{3} \][/tex]
9. Find Corresponding [tex]\( x \)[/tex] Values:
For [tex]\( y = 2 + \sqrt{3} \)[/tex]:
[tex]\[ x = 5 - 2(2 + \sqrt{3}) = 5 - 4 - 2\sqrt{3} = 1 - 2\sqrt{3} \][/tex]
For [tex]\( y = 2 - \sqrt{3} \)[/tex]:
[tex]\[ x = 5 - 2(2 - \sqrt{3}) = 5 - 4 + 2\sqrt{3} = 1 + 2\sqrt{3} \][/tex]
10. Points of Intersection:
The points of intersection (solutions) are:
[tex]\[ \left(1 - 2\sqrt{3}, 2 + \sqrt{3}\right) \quad \text{and} \quad \left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right) \][/tex]
11. Identify Point B:
Since point A is [tex]\((-7, 6)\)[/tex] and it is not either of these solutions, point B must be [tex]\(\left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right)\)[/tex].
Therefore, the coordinates of point B are:
[tex]\[ \boxed{\left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right)} \][/tex]
1. Equation of the Circle:
The circle is centered at the origin and has the equation:
[tex]\[ x^2 + y^2 = 20 \][/tex]
2. Equation of the Line:
The line has the equation:
[tex]\[ x + 2y = 5 \][/tex]
3. Solving the System of Equations:
We solve these two equations simultaneously to find the points of intersection (where the line intersects the circle).
4. Substitution Method:
From the line equation [tex]\( x + 2y = 5 \)[/tex], solve for [tex]\( x \)[/tex]:
[tex]\[ x = 5 - 2y \][/tex]
5. Substitute into the Circle Equation:
Substitute [tex]\( x = 5 - 2y \)[/tex] into the circle equation [tex]\( x^2 + y^2 = 20 \)[/tex]:
[tex]\[ (5 - 2y)^2 + y^2 = 20 \][/tex]
6. Expand and Simplify:
[tex]\[ (5 - 2y)^2 = 25 - 20y + 4y^2 \][/tex]
So, the equation becomes:
[tex]\[ 25 - 20y + 4y^2 + y^2 = 20 \][/tex]
Combine like terms:
[tex]\[ 5y^2 - 20y + 25 = 20 \][/tex]
Subtract 20 from both sides:
[tex]\[ 5y^2 - 20y + 5 = 0 \][/tex]
7. Simplify the Quadratic Equation:
[tex]\[ y^2 - 4y + 1 = 0 \][/tex]
8. Solve the Quadratic Equation (Use the Quadratic Formula):
Recall, the quadratic formula is [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex].
For our quadratic equation [tex]\( y^2 - 4y + 1 = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ y = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ y = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ y = 2 \pm \sqrt{3} \][/tex]
9. Find Corresponding [tex]\( x \)[/tex] Values:
For [tex]\( y = 2 + \sqrt{3} \)[/tex]:
[tex]\[ x = 5 - 2(2 + \sqrt{3}) = 5 - 4 - 2\sqrt{3} = 1 - 2\sqrt{3} \][/tex]
For [tex]\( y = 2 - \sqrt{3} \)[/tex]:
[tex]\[ x = 5 - 2(2 - \sqrt{3}) = 5 - 4 + 2\sqrt{3} = 1 + 2\sqrt{3} \][/tex]
10. Points of Intersection:
The points of intersection (solutions) are:
[tex]\[ \left(1 - 2\sqrt{3}, 2 + \sqrt{3}\right) \quad \text{and} \quad \left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right) \][/tex]
11. Identify Point B:
Since point A is [tex]\((-7, 6)\)[/tex] and it is not either of these solutions, point B must be [tex]\(\left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right)\)[/tex].
Therefore, the coordinates of point B are:
[tex]\[ \boxed{\left(1 + 2\sqrt{3}, 2 - \sqrt{3}\right)} \][/tex]
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