A tangent is a line that touches a circle at exactly one point. For what values of k will the line y= x+k be tangent to the circle x^2+ y^2 = 25?
I know I'm suppose to use the properties of a discrimminate. But how
the intersection(s) will be where
x^2 + (x+k)^2 - 25 has one solution. That is, where the discriminant is zero.
x^2 + x^2 + 2kx + k^2-25 = 0
2x^2 + 2kx + (k^2-25) = 0
The discriminant is
(2k)^2 - 4(2)(k^2-25)
4k^2 - 8k^2 + 200 = 0
k^2 = 50
k = ±√50
So, check out the graphs at
http://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%3D25%2C+y%3Dx%2B%E2%88%9A50%2C+y%3Dx-%E2%88%9A50
Well, let's see if we can unravel this tangential situation and have some fun with it!
So, we have a circle with the equation x^2 + y^2 = 25, and we want to find the values of k that make the line y = x + k tangent to this circle.
To do that, we need to find where the line and the circle intersect. Since the line is tangent to the circle, there should be exactly one point of intersection.
Substituting y = x + k into the equation of the circle, we get x^2 + (x + k)^2 = 25.
Expanding this equation, we have x^2 + (x^2 + 2kx + k^2) = 25.
Combining like terms, we have 2x^2 + 2kx + k^2 = 25.
Now, this is a quadratic equation, and we can use the discriminant to see when the line and the circle intersect at exactly one point.
The discriminant is given by b^2 - 4ac, where a = 2, b = 2k, and c = k^2 - 25.
In order to have exactly one point of intersection, the discriminant must be equal to zero.
So we have (2k)^2 - 4(2)(k^2 - 25) = 0.
Simplifying this equation, we get 4k^2 - 8k^2 + 200 = 0.
Combining like terms, we have -4k^2 + 200 = 0.
Dividing both sides of the equation by -4, we obtain k^2 - 50 = 0.
Now, solving for k in this equation, we find that k = ±√50.
So, the line y = x + k is tangent to the circle x^2 + y^2 = 25 when k is equal to ±√50.
And that's how the tangential tale of the line and the circle comes to a hilariously satisfying conclusion!
To find the values of k for which the line y = x + k is tangent to the circle x^2 + y^2 = 25, we can substitute y = x + k into the equation of the circle and solve for x.
Substituting y = x + k into the equation x^2 + y^2 = 25, we get:
x^2 + (x + k)^2 = 25
Expanding and simplifying this equation gives us:
x^2 + x^2 + 2kx + k^2 = 25
2x^2 + 2kx + k^2 = 25
Now we have a quadratic equation in terms of x. To find the values of x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = 2k, and c = k^2 - 25.
Substituting these values into the quadratic formula, we get:
x = (-2k ± √((2k)^2 - 4(2)(k^2 - 25))) / (2(2))
x = (-2k ± √(4k^2 - 8k^2 + 200)) / 4
x = (-2k ± √(-4k^2 + 200)) / 4
x = (-2k ± √(4(50 - k^2))) / 4
x = (-2k ± 2√(50 - k^2)) / 4
x = -k ± √(50 - k^2) / 2
Since the line is tangent to the circle, there will be exactly one solution for x. This means that the discriminant of the quadratic equation, (50 - k^2), must be equal to zero.
Setting the discriminant equal to zero, we get:
50 - k^2 = 0
k^2 = 50
Taking the square root of both sides, we have two possible values for k:
k = √50
k = -√50
Therefore, the values of k for which the line y = x + k is tangent to the circle x^2 + y^2 = 25 are k = √50 and k = -√50.
To find the values of k for which the line y = x + k is tangent to the circle x^2 + y^2 = 25, we need to determine the conditions under which the line and the circle have exactly one point of intersection.
The equation of the circle is:
x^2 + y^2 = 25
Substitute y = x + k (the equation of the line) into the circle equation:
x^2 + (x + k)^2 = 25
Expand and simplify the equation:
x^2 + (x^2 + 2kx + k^2) = 25
2x^2 + 2kx + k^2 - 25 = 0
Now, we can compare this quadratic equation with the general form of a quadratic equation ax^2 + bx + c = 0. In this case, the quadratic equation will have exactly one solution if its discriminant (b^2 - 4ac) is equal to zero.
The discriminant of the quadratic equation is:
D = (2k)^2 - 4(2)(k^2 - 25)
D = 4k^2 - 8(k^2 - 25)
D = 4k^2 - 8k^2 + 200
D = -4k^2 + 200
For the line to be tangent to the circle, the discriminant D must be zero:
-4k^2 + 200 = 0
Rearrange the equation to solve for k:
4k^2 = 200
k^2 = 50
k = ±√50
So, the line y = x + k is tangent to the circle x^2 + y^2 = 25 for k = √50 and k = -√50.