Find the equations of the tangent from the origin to the circle
x^2+y^2-5x-5y+10=0
differentiate implicitly ...
2x + 2y dy/dx - 5 - 5dy/dx = 0
dy/dx (2y - 5) = 5 - 2x
dy/dx = (5-2x)/(2y-5)
let the point of contact be (a,b)
slope of tangent = (5-2a)/(2b-5)
but the slope of the tangent from (a,b) to (0,0) = b/a
so (5-2a)/(2b-5) = b/a
5a - 2a^2 = 2b^2 - 5b
2a^2 + 2b^2 = 5a + 5b
2(a^2+b^2) = 5(a+b)
a^2 + b^2 = (5/2)(a+b)
but since (a,b) lies on the curve,
a^2 + b^2 - 5a - 5b + 10 = 0
sub in the above
(5/2)(a+b) - 5a - 5b =- 10
times 2
5(a+b) - 10a - 10b = -20
-5a-5b = -20
a+b = 4
a = 4-b
sub into a^2 + b^2 = (5/2)(a+b)
(4-b)^2 + b^2 = (5/2)(4-b + b)
b^2 - 8b + 16 + b^2 = 10
2b^2 - 8b + 6 = 0
b^2 - 4b + 3 = 0
(b-1)(b-3) = 0
b = 1 or b = 3
then a = 3 or a = 1
There will be 2 points of contact, (1,3) and (3,1)
I will do one of the tangents, you do the other one
1. for the point (1,3) and the origin,
slope of tangent = 3/1 = 3
equation of tangent is y = 3x
Why did the circle go to the therapist? It had some tangent issues!
To find the equations of the tangents from the origin to the circle, we first need to determine the point(s) where the tangents touch the circle.
We start by rewriting the equation of the circle in the standard form:
x^2 + y^2 - 5x - 5y + 10 = 0
Completing the squares, we get:
(x^2 - 5x + 25/4) + (y^2 - 5y + 25/4) = 10 + 25/4
(x - 5/2)^2 + (y - 5/2)^2 = 15/4
Now, we know that the origin is (0, 0). We substitute these values into the equation of the circle:
(0 - 5/2)^2 + (0 - 5/2)^2 = 15/4
(25/4) + (25/4) = 15/4
50/4 = 15/4
Since this equation is not true, the origin does not lie on the circle. Hence, there are no tangents from the origin to this circle.
But chin up, there are plenty of other circles out there that would be more than happy to be tangent buddies with the origin!
To find the equations of the tangents from the origin to the circle x^2 + y^2 - 5x - 5y + 10 = 0, we need to follow these steps:
Step 1: Calculate the center and radius of the circle.
Step 2: Find the slope of the line connecting the origin to the center of the circle.
Step 3: Find the slopes of the tangents from the origin to the circle.
Step 4: Use the point-slope form of a line to find the equations of the tangents.
Let's calculate these steps one by one.
Step 1: Calculate the center and radius of the circle.
To find the center and radius of the circle, we need to rewrite the equation in the standard form, (x - h)^2 + (y - k)^2 = r^2.
Rearranging the given equation: x^2 + y^2 - 5x - 5y + 10 = 0
(x^2 - 5x) + (y^2 - 5y) + 10 = 0
Completing the square for x terms: (x^2 - 5x + 25/4) + (y^2 - 5y) + 10 = 25/4
Completing the square for y terms: (x^2 - 5x + 25/4) + (y^2 - 5y + 25/4) + (10 - 25/4) = 25/4
(x - 5/2)^2 + (y - 5/2)^2 = 15/4
So, the center of the circle is (5/2, 5/2) and the radius is √(15/4) = √15/2.
Step 2: Find the slope of the line connecting the origin to the center of the circle.
The slope of the line connecting the origin (0, 0) to the center (5/2, 5/2) is given by the formula: slope = (y2 - y1) / (x2 - x1).
Substituting the values: slope = (5/2 - 0) / (5/2 - 0) = 5/2 / 5/2 = 1.
Step 3: Find the slopes of the tangents from the origin to the circle.
The slopes of the tangents from the origin to the circle can be found using the formula: slope = -1 / m, where m is the slope of the line connecting the origin to the center.
Substituting the value: slope of the tangents = -1 / 1 = -1.
Step 4: Use the point-slope form of a line to find the equations of the tangents.
The point-slope form of a line is given by the equation: y - y1 = m(x - x1), where (x1, y1) is a point on the line.
For the first tangent, we have the point (0, 0) on the line. Using this in the equation, we get: y - 0 = -1(x - 0).
Simplifying, we have y = -x.
For the second tangent, we also have the point (0, 0) on the line. Using this in the equation, we get: y - 0 = -1(x - 0).
Again, we have y = -x.
Therefore, the equations of the tangents from the origin to the circle x^2 + y^2 - 5x - 5y + 10 = 0 are y = -x.
To find the equation of the tangent line from the origin to the given circle, we need to find the point of tangency first. Here's how you can do it:
Step 1: Rewrite the equation of the circle in standard form.
x^2 + y^2 - 5x - 5y + 10 = 0
Rearrange the equation:
x^2 - 5x + y^2 - 5y = -10
Step 2: Complete the square for both the x and y terms.
For x terms:
x^2 - 5x = -(y^2 - 5y + 10)
To complete the square, take half of the coefficient of the middle term (-5) and square it.
(-5/2)^2 = 25/4
To maintain the equation's balance, we need to add 25/4 on both sides of the equation:
x^2 - 5x + 25/4 = -(y^2 - 5y + 10) + 25/4
Now, the left side can be factored as a perfect square: (x - 5/2)^2
And, the right side simplifies to: -(y^2 - 5y + 35/4)
The equation becomes:
(x - 5/2)^2 = -(y^2 - 5y + 35/4)
Step 3: Convert the equation to the standard form.
(x - 5/2)^2 = -[(y^2 - 5y + 35/4)]
Multiply both sides by -1 to remove the negative sign:
-(x - 5/2)^2 = y^2 - 5y + 35/4
Rearrange the equation:
(x - 5/2)^2 + (y^2 - 5y + 35/4) = 0
Step 4: Identify the center and radius of the circle.
The equation is now in the form (x - h)^2 + (y - k)^2 = r^2.
The center, (h, k), is (5/2, 5/2), and the radius squared, r^2, is 0.
Step 5: Find the point of tangency.
Since the radius is 0, it means the circle is degenerate, and it forms a single point.
Therefore, the point of tangency is just the center of the circle, which is (5/2, 5/2).
Step 6: Find the equation of the tangent line.
Since the tangent line passes through the origin and the point of tangency, we can use the two-point form of a line.
Let the equation of the tangent line be y = mx, where m is the slope we need to find.
The slope, m, can be found using the point-slope form:
m = (y2 - y1)/(x2 - x1)
Plugging in the coordinates of the center and the origin:
m = (5/2 - 0) / (5/2 - 0)
m = (5/2) / (5/2)
m = 1
Now, we can write the equation of the tangent line:
y = mx
y = x
Therefore, the equation of the tangent line from the origin to the given circle x^2 + y^2 - 5x - 5y + 10 = 0 is y = x.