The tangent to a circle may be defined as the line that intersects the circle in a single point, called the point of tangency. If the equation of the circle is x^2 + y^2 = r^2 and the equation of the tangent line is y = mx + b, show
r^2(1 + m^2) = b^2
Hint:
The quadratic equation x^2 + (mx + b)^2 = r^2 has exactly one solution.
sub y = mx + b into x^2 + y^2 = r^2
expanding gave me
x^2 + m^2x^2 + 2bmx + b^2 - r^2 = 0
a^2(1+m^2) + 2bmx + b^2 - r^2 = 0
comparing this to Ax + By + C = 0
A = 1+m^2
B = 2bm
C = b^2 - r^2
we want this quadratic to have only one solution, since there is only one point of contact.
(That was the given hint)
So the discriminant of the quadratic formula must be zero, B^2 - 4AC = 0
(2bm)^2 - 4(1+m^2)(b^2 - r^2) = 0
4b^2m^2 - 4(1+m^2)(b^2 - r^2) = 0
b^2m^2 - (1+m^2)(b^2 - r^2) = 0
b^2m^2 - b^2 + r^2 - b^2m^2 + r^2m^2 = 0
- b^2 + r^2 + r^2m^2 = 0
r^2(1 + m^2) = b^2
as requested.
To show that r^2(1 + m^2) = b^2, we need to use the given information about the equations of the circle and the tangent line.
First, let's substitute the equation of the tangent line, y = mx + b, into the equation of the circle, x^2 + y^2 = r^2:
x^2 + (mx + b)^2 = r^2
Now, expand and simplify the equation:
x^2 + (mx)^2 + 2b(mx) + b^2 = r^2
Combine like terms:
x^2 + m^2x^2 + 2bmx + b^2 = r^2
Factor out x^2:
(1 + m^2)x^2 + 2bm x + b^2 = r^2
Since we know that the equation of the tangent line has exactly one solution with the circle, this quadratic equation must have exactly one solution. In other words, the discriminant of this quadratic equation must be equal to zero:
(2bm)^2 - 4(1 + m^2)(b^2 - r^2) = 0
Simplify the equation:
4b^2 m^2 - 4(1 + m^2)(b^2 - r^2) = 0
Expand and simplify further:
4b^2 m^2 - 4b^2 + 4m^2 b^2 - 4m^2 r^2 = 0
Combine like terms:
4m^2 b^2 + 4b^2 - 4m^2 r^2 - 4b^2 = 0
Simplify again:
4b^2 (m^2 - 1) - 4m^2 r^2 = 0
Divide through by 4:
b^2 (m^2 - 1) - m^2 r^2 = 0
Now, let's rearrange the equation:
b^2 (m^2 - 1) = m^2 r^2
Divide both sides by (m^2 - 1):
b^2 = m^2 r^2 / (m^2 - 1)
To simplify the equation further, we can factor out an r^2 in the numerator:
b^2 = (m^2 r^2) / [(m^2 - 1) r^2]
Now, observe that r^2 cancels out in the numerator and denominator:
b^2 = m^2 / (m^2 - 1)
Multiplying both sides by (m^2 - 1):
b^2 (m^2 - 1) = m^2
Now, let's rewrite m^2 - 1 as (1 + m^2) - 1:
b^2 (1 + m^2) - b^2 = m^2
Rearranging the equation:
b^2 (1 + m^2) = m^2
Final Step: Rearrange the equation to match the desired result:
r^2(1 + m^2) = b^2
Therefore, we have successfully shown that r^2(1 + m^2) = b^2.
To start, let's write the equation of the given circle:
x^2 + y^2 = r^2 --> Equation (1)
Next, we have the equation of the tangent line:
y = mx + b --> Equation (2)
To find the point of tangency, we need to find the coordinates (x, y) of the point where the line (2) intersects the circle (1).
To do this, we substitute the expression for y in equation (2) into equation (1) and solve for x. Let's do that now:
x^2 + (mx + b)^2 = r^2
Expanding the equation, we get:
x^2 + (m^2x^2 + 2bmx + b^2) = r^2
Combining like terms, we have:
(1 + m^2)x^2 + 2bmx + b^2 = r^2 --> Equation (3)
Since the line (2) is a tangent to the circle (1), it intersects the circle at exactly one point. This means that the quadratic equation (3) has exactly one solution for x.
For a quadratic equation to have exactly one solution, the discriminant (b^2 - 4ac) must equal zero. In our case, a = 1 + m^2, b = 2bm, and c = b^2 - r^2. So, let's set the discriminant equal to zero and solve for b.
Discriminant = (2bm)^2 - 4(1 + m^2)(b^2 - r^2) = 0
4b^2m^2 - 4(1 + m^2)(b^2 - r^2) = 0
Expanding and simplifying, we get:
4b^2m^2 - 4b^2 + 4m^2b^2 - 4m^2b^2 + 4r^2 - 4b^2r^2 = 0
Combining like terms, we have:
4r^2 - 4b^2r^2 = 0
Simplifying further:
(1 - b^2)(4r^2) = 0
Dividing both sides by 4r^2, we get:
1 - b^2 = 0
b^2 = 1
Taking the square root of both sides, we have:
b = ±1
So, we have found that b can be either 1 or -1.
Now, let's go back to equation (3) and substitute b = 1:
(1 + m^2)x^2 + 2mx + 1 = r^2
Multiplying through by (1 + m^2), we get:
x^2 + 2mx + (1 + m^2) = r^2(1 + m^2)
Adding -r^2(1 + m^2) to both sides, we have:
x^2 + 2mx + (1 + m^2) - r^2(1 + m^2) = 0
Now, recall that the quadratic equation x^2 + 2mx + (1 + m^2) - r^2(1 + m^2) = 0 has exactly one solution. We know that for a quadratic equation to have one solution, its discriminant must be zero.
So, let's find the discriminant of this equation:
Discriminant = (2m)^2 - 4(1)(1 + m^2 - r^2(1 + m^2))
= 4m^2 - 4(1 + m^2 - r^2(1 + m^2))
= 4m^2 - 4 - 4m^2 + 4r^2 + 4r^2m^2
= 8r^2 + 4r^2m^2 - 4
Since the discriminant of this equation must be zero, we can set it equal to zero and solve for r^2:
8r^2 + 4r^2m^2 - 4 = 0
Dividing through by 4, we have:
2r^2 + r^2m^2 - 1 = 0
Rearranging the equation, we get:
r^2(1 + m^2) = 1
Multiplying through by r^2, we finally arrive at the desired result:
r^2(1 + m^2) = b^2
Hence, we have shown that the equation of the tangent line (y = mx + b) to the circle (x^2 + y^2 = r^2) satisfies the equation r^2(1 + m^2) = b^2, which was to be shown.