Let the line X-2Y=15 and the circle with equation X^2+Y^2=50 intersect at the points A and B.
Algebraically find the coordinates of A and B using the information given.
x=2y+15
(2y+15)^2 + y^2=50
expand that out, solve for y. You may have to use the quadratic equation to find the two y solutions. Then put each back into either equation, and find the two x solutions.
Sure, here we go!
First, let's rewrite the equation of the line in terms of Y:
X - 2Y = 15
2Y = X - 15
Y = (1/2)X - (15/2)
Now we substitute this expression for Y into the equation of the circle to find the x-coordinates of the points of intersection:
X^2 + (1/2)X - (15/2)^2 = 50
Simplifying:
X^2 + (1/2)X - 225/4 = 200/4
X^2 + (1/2)X - 225/4 = 50
To make things simpler, let's multiply everything by 4 to clear the fractions:
4X^2 + 2X - 225 = 200
Rearranging the equation:
4X^2 + 2X - 425 = 0
Now we can solve this quadratic equation using the quadratic formula:
X = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
X = (-2 ± √(2^2 - 4(4)(-425))) / (2(4))
Calculating the discriminant under the square root:
X = (-2 ± √(4 + 6800)) / 8
X = (-2 ± √6804) / 8
X = (-2 ± 82.5) / 8
Simplifying:
X1 = (-2 + 82.5) / 8
X1 = 80.5 / 8
X1 = 10.0625
X2 = (-2 - 82.5) / 8
X2 = -84.5 / 8
X2 = -10.5625
Now we substitute these values back into the equation of the line to find the corresponding y-coordinates:
For X1 = 10.0625:
Y1 = (1/2)(10.0625) - 15/2
Y1 = 5.03125 - 7.5
Y1 = -2.46875
For X2 = -10.5625:
Y2 = (1/2)(-10.5625) - 15/2
Y2 = -5.28125 - 7.5
Y2 = -12.78125
So, the coordinates of point A are (10.0625, -2.46875), and the coordinates of point B are (-10.5625, -12.78125). Enjoy the intersection, it's a great place where lines and circles meet and discuss their differences!
To find the coordinates of the points A and B where the line X-2Y=15 and the circle X^2+Y^2=50 intersect, we need to solve these two equations simultaneously.
Step 1: Solve the linear equation for X in terms of Y.
X - 2Y = 15
X = 2Y + 15
Step 2: Substitute the expression for X in the equation of the circle.
(2Y + 15)^2 + Y^2 = 50
4Y^2 + 60Y + 225 + Y^2 = 100
5Y^2 + 60Y + 125 = 100
Step 3: Simplify the equation by subtracting 100 from both sides.
5Y^2 + 60Y + 25 = 0
Step 4: Factor the quadratic equation.
(5Y + 5)(Y + 5) = 0
Step 5: Set each factor equal to 0 and solve for Y.
5Y + 5 = 0 or Y + 5 = 0
5Y = -5 or Y = -5
Y = -1 or Y = -5
Step 6: Substitute the values of Y back into the equation X = 2Y + 15 to find the corresponding X-values.
For Y = -1, X = 2(-1) + 15 = 13.
So the coordinates of point A are (13, -1).
For Y = -5, X = 2(-5) + 15 = 5.
So the coordinates of point B are (5, -5).
Therefore, the coordinates of points A and B are (13, -1) and (5, -5), respectively.
To find the coordinates of the points A and B where the line and the circle intersect, we need to solve the system of equations formed by the line equation X-2Y=15 and the circle equation X^2+Y^2=50.
1. Solve the linear equation X-2Y=15 for one variable, X or Y, in terms of the other variable. Let's solve it for X:
X = 15 + 2Y
2. Substitute this value of X into the circle equation:
(15 + 2Y)^2 + Y^2 = 50
3. Expand and simplify the equation:
225 + 60Y + 4Y^2 + Y^2 = 50
5Y^2 + 60Y + 175 = 50
4. Subtract 50 from both sides of the equation:
5Y^2 + 60Y + 125 = 0
5. Divide the entire equation by 5 to simplify it further:
Y^2 + 12Y + 25 = 0
6. Factor the quadratic equation:
(Y + 5)(Y + 5) = 0
7. Set each factor equal to zero:
Y + 5 = 0
Solving for Y, we get:
Y = -5
8. Substitute this value of Y back into the linear equation to find X:
X = 15 + 2(-5)
X = 15 - 10
X = 5
9. The coordinates of one intersection point, which we'll call A, are (5, -5).
Now let's find the other intersection point, B.
10. Substitute Y = -5 into the linear equation to find X again:
X = 15 + 2(-5)
X = 15 - 10
X = 5
11. The coordinates of the other intersection point, which we'll call B, are also (5, -5).
So, the coordinates of both points A and B are (5, -5).