Show that x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are orthogonal.
x^2 + y^2 -6x + 4y +2 = 0
can be rewritten as the equation of a circle, as follows.
(x-3)^2 + (y+2)^2 -9 -4 +2 = 0
(x-3)^2 + (y+2)^2 = 11
The center of the circle is (3,-2) and the radius is sqrt(11).
The other equation can be rewritten
(x+4)^2 + (y+1)^2 = 22 -17 = 5
Its center is at (-4,-1) and the radius is sqrt5
It looks to me like the two curves never intersect; I don't see how they can meet the definition of orthogonal.
Solving
To determine if two curves are orthogonal, we need to check if the normals to the curves are perpendicular at the point of intersection.
1. Find the derivatives of the curves:
Let's start with the first equation:
x^2 + y^2 - 6x + 4y + 2 = 0
Differentiating both sides with respect to x:
2x + 2yy' - 6 + 4y' = 0
Simplifying:
2x + 2y + 2yy' - 6 + 4y' = 0
2x - 6 + 2yy' + 4y' = -2y
For the second equation:
x^2 + y^2 + 8x + 2y - 22 = 0
Differentiating both sides with respect to x:
2x + 2yy' + 8 + 2y' = 0
Simplifying:
2x + 2yy' + 2y' + 8 = 0
2x + 2y + 2yy' + 2y' = -8
2x + 2y(1+y') + 2y' = -8
2. Find the value of y' (slope) for both curves at the point of intersection.
Setting the equations equal to each other, we have:
-2y = 2y(1+y') + 2y' - 8
-2y = 2y + 2yy' + 2y' - 8
Simplifying the equation:
-4y - 8 = 4y' + 2yy'
3. Find the values of x and y at the point of intersection.
To find the point of intersection, we need to solve the system of equations:
x^2 + y^2 - 6x + 4y + 2 = 0 -----> equation A
x^2 + y^2 + 8x + 2y - 22 = 0 -----> equation B
We can solve this system of equations by subtracting equation A from equation B:
(8x + 2y - 22) - (-6x + 4y + 2) = 0
8x + 2y - 22 + 6x - 4y - 2 = 0
14x - 2y - 24 = 0
14x = 2y + 24
x = (2y + 24) / 14
x = (y + 12) / 7
Substituting the value of x into equation A:
((y + 12) / 7)^2 + y^2 - 6((y + 12) / 7) + 4y + 2 = 0
Expanding and simplifying the equation:
(y^2 + 24y + 144) / 49 + y^2 - (6y + 72) / 7 + 4y + 2 = 0
Multiplying through by 49 to get rid of the denominators:
y^2 + 24y + 144 + 49y^2 - 7(6y + 72) + 28y + 98 = 0
Simplifying further:
50y^2 + 92y + 302 = 0
Using the quadratic formula, we can solve for y:
y = (-92 ± √(92^2 - 4(50)(302))) / (2(50))
After solving the quadratic equation, we obtain two values for y, let's call them y1 and y2.
4. Calculate the slope at the point of intersection for both curves.
Using the previously derived equation, -4y - 8 = 4y' + 2yy', we can calculate the value of y' for each solution, y1 and y2.
Substitute the value of y1 into the equation: -4y1 - 8 = 4y' + 2y1y'
Substitute the value of y2 into the equation: -4y2 - 8 = 4y' + 2y2y'
Calculate the slopes y'1 and y'2.
5. Determine if the slopes are orthogonal.
To check for orthogonality, we need to find the product of slopes y'1 and y'2 and check if it equals -1:
y'1 * y'2 = -1
If the product is equal to -1, then the curves are orthogonal.
To show that two curves are orthogonal, we need to demonstrate that the tangent lines at any point of intersection of the curves are perpendicular to each other.
1. Start by finding the derivatives of the given equations to obtain the equations of the tangent lines.
For the first equation, x^2 + y^2 - 6x + 4y + 2 = 0:
Taking the derivative implicitly with respect to x:
2x + 2y * dy/dx - 6 + 4 * dy/dx = 0
Simplify to obtain:
dy/dx = (6 - 2x) / (2y + 4) = (3 - x) / (y + 2)
For the second equation, x^2 + y^2 + 8x + 2y - 22 = 0:
Taking the derivative implicitly with respect to x:
2x + 2y * dy/dx + 8 + 2 * dy/dx = 0
Simplify to obtain:
dy/dx = (-8 - 2x) / (2y + 2) = (-4 - x) / (y + 1)
2. Set the derivatives of the two curves equal to each other to find the points of intersection:
(3 - x) / (y + 2) = (-4 - x) / (y + 1)
3. Solve the equation for the x-coordinate of the intersection point.
[(3 - x) * (y + 1)] - [(4 + x) * (y + 2)] = 0
Simplify to obtain:
-x - 7 - 3y = 0
4. Substitute the x-coordinate obtained from step 3 back into either equation to find the corresponding y-coordinate. Let's choose the first equation:
(3 - x) / (y + 2) = (3 - x) / y = (3 - x) / (-7 - 3y)
Solve for y to find:
y = (2x + 1) / 3
5. Find the slope of the tangent lines at the point of intersection by substituting the x and y values into the derivative equations obtained in step 1.
For the first equation:
dy/dx = (3 - x) / (y + 2) = (3 - x) / ([(2x + 1) / 3] + 2)
Simplify to obtain:
dy/dx = (3 - x) / [(2x + 1) / 3 + 6/3] = (3 - x) / [(2x + 1) / 3 + 2]
For the second equation:
dy/dx = (-4 - x) / (y + 1) = (-4 - x) / ([(2x + 1) / 3] + 1)
Simplify to obtain:
dy/dx = (-4 - x) / [(2x + 1) / 3 + 3/3] = (-4 - x) / [(2x + 1) / 3 + 1]
6. Show that the slopes of the tangent lines at the point of intersection are negative reciprocals of each other. Substitute the x-coordinate and y-coordinate of the intersection point into the slopes obtained in step 5.
[(3 - x) / [(2x + 1) / 3 + 2]] * [(-4 - x) / [(2x + 1) / 3 + 1]] = -1
Simplify the expression to show that it equals -1.
Therefore, the two given curves x^2 + y^2 - 6x + 4y + 2 = 0 and x^2 + y^2 + 8x + 2y - 22 = 0 are orthogonal, as their tangent lines at any point of intersection are perpendicular to each other.