show that the circle x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are orthogonal
The circles are
(x-3)^2 + (y+2)^2 = 11
(x+4)^2 + (y+1)^2 = 39
So their centers are at (3,-2) and (-4,-1)
The circles intersect at
(73/50 ± √429/50 , -89/50 ± 7√429/50)
or
(1.046,-4.680) and (1.874,1.120)
Now, using those two points, and knowing the centers of the circles, you can show that the lines joining the centers to the intersections are perpendicular.
Since the radii are perpendicular, so are the tangents, and thus the circles.
Solve that question
Sols
To determine if two circles are orthogonal, we need to check if the common tangents of the circles are mutually perpendicular.
Step 1: Rewrite the equations of the circles in standard form.
The equation of the first circle is: x^2 + y^2 - 6x + 4y + 2 = 0.
Rearranging terms, we get: (x^2 - 6x) + (y^2 + 4y) = -2.
Completing the square, we have: (x^2 - 6x + 9) + (y^2 + 4y + 4) = -2 + 9 + 4.
Simplifying further, we get: (x - 3)^2 + (y + 2)^2 = 11.
The equation of the second circle is: x^2 + y^2 + 8x + 2y - 22 = 0.
Rearranging terms, we get: (x^2 + 8x) + (y^2 + 2y) = 22.
Completing the square, we have: (x^2 + 8x + 16) + (y^2 + 2y + 1) = 22 + 16 + 1.
Simplifying further, we get: (x + 4)^2 + (y + 1)^2 = 39.
Step 2: Find the centers and radii of the circles.
For the first circle, the center is (3, -2) and the radius is sqrt(11).
For the second circle, the center is (-4, -1) and the radius is sqrt(39).
Step 3: Determine the slopes of the common tangents.
To find the slopes of the common tangents, we can use the fact that a line passing through the center of a circle is perpendicular to the tangent at the point of contact.
The slope of a line passing through the center (h, k) of a circle with equation (x - h)^2 + (y - k)^2 = r^2 is given by -h/k.
For the first circle, the center is (3, -2), so the slope of the first common tangent is -3/-2 = 3/2.
For the second circle, the center is (-4, -1), so the slope of the second common tangent is -(-4)/(-1) = 4.
Step 4: Check if the slopes of the common tangents are mutually perpendicular.
The slopes 3/2 and 4 are not mutually perpendicular (their product is not -1). Therefore, the circles x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are not orthogonal.
To determine if two circles are orthogonal (perpendicular), we need to check if the tangent lines at their points of intersection are perpendicular.
First, let's find the points of intersection between the two circles. We can do this by solving their respective equations simultaneously.
The first circle is given by:
x^2 + y^2 - 6x + 4y + 2 = 0
The second circle is given by:
x^2 + y^2 + 8x + 2y - 22 = 0
To find the points of intersection, we need to eliminate one variable. Let's eliminate y by subtracting the equations:
(x^2 + y^2 + 8x + 2y - 22) - (x^2 + y^2 - 6x + 4y + 2) = 0
Simplifying the equation, we get:
14x - 2y - 24 = 0
Now, let's solve this equation for y:
-2y = -14x + 24
y = 7x - 12
Now, substitute this value of y into any of the circle equations to solve for x. Let's use the first circle equation:
x^2 + (7x - 12)^2 - 6x + 4(7x - 12) + 2 = 0
Expanding and rearranging, we get:
x^2 + 49x^2 - 168x + 144 - 6x + 28x - 48 + 2 = 0
50x^2 - 146x + 98 = 0
Now, solve this quadratic equation to find the values of x. We can do this by factoring or using the quadratic formula.
By solving the equation, we find two distinct values for x: x = 0.5882 and x = 1.176.
Now, substitute these x-values back into the equation y = 7x - 12 to find the corresponding y-values.
At x = 0.5882, y = 7(0.5882) - 12 ≈ -7.065
At x = 1.176, y = 7(1.176) - 12 ≈ -6.228
We have found the two points of intersection: (0.5882, -7.065) and (1.176, -6.228).
Next, let's find the slopes of the tangent lines at these points on each circle. The slope of a tangent line to a circle at a given point (x, y) is given by the negative reciprocal of the derivative of the circle equation at that point.
For the first circle:
x^2 + y^2 - 6x + 4y + 2 = 0
Taking the derivative with respect to x, we get:
2x - 6 + 4y dy/dx + 4 = 0
dy/dx = (6 - 2x) / (4 + 4y)
At the point (0.5882, -7.065), the slope of the first circle is:
dy/dx = (6 - 2(0.5882)) / (4 + 4(-7.065)) ≈ 0.5
For the second circle:
x^2 + y^2 + 8x + 2y - 22 = 0
Taking the derivative with respect to x, we get:
2x + 2y dy/dx + 8 = 0
dy/dx = (-2x - 8) / (2y)
At the point (0.5882, -7.065), the slope of the second circle is:
dy/dx = (-2(0.5882) - 8) / (2(-7.065)) ≈ -0.5
Now, let's calculate the product of the slopes at the point of intersection:
0.5 * (-0.5) = -0.25
Since the product of the slopes is -0.25, which is equal to -1/4, we can conclude that the tangent lines at the points of intersection are perpendicular. Therefore, the two circles x^2+y^2-6x+4y+2=0 and x^2+y^2+8x+2y-22=0 are orthogonal.