Hi, I am studying for a precalculus quiz and I do not understand this hw problem:
"The tangent line to a circle may be defined as the point that intersects a circle in a single point... 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 that:
a. r^2(1+m^2)=b^2 [Hint: the quadratic equation x^2+(mx+b)^2=r^2 has exactly one solution]
b.the point of tangency is (-r^2m/b,r^2/b)
c. the tangent line is perpendicular to the line containing the circle and the point of tangency
a. For the equation
x^2+(mx+b)^2 = r^2
to have one solution, the discriminant of the quadratic equation
Ax^2 + Bx + C = 0
must be zero
B^2 = 4 AC
A = 1+m^2
B = 2mb
C = b^2-r^2
4m^2b^2 = 4(1+m^2)(b^2-r^2)
m^2b^2 = b^2 +b^2m^2 -r^2 -r^2m^2
b^2 -r^2(1+m^2) = 0
b. x = -B/(2A) = -mb/(1+m^2)
= -mb[r^2/b^2] = -mr^2/b
c. Don't you mean the line containing the CENTER OF the circle and the point of tangency?
Thank you so much. you totally rock
To understand and solve this problem, let's break it down step by step:
a. To show that r^2(1+m^2) = b^2, we need to prove that the quadratic equation x^2 + (mx + b)^2 = r^2 has exactly one solution.
1. Start by substituting the equation of the circle, x^2 + y^2 = r^2, into the quadratic equation.
x^2 + (mx + b)^2 = r^2
2. Simplify the equation by expanding the squared term.
x^2 + (m^2x^2 + 2mxb + b^2) = r^2
3. Combine like terms and rewrite the equation as a quadratic equation.
(1 + m^2)x^2 + 2mbx + (b^2 - r^2) = 0
4. For a quadratic equation to have exactly one solution, its discriminant (b^2 - 4ac) must be equal to zero.
5. The discriminant of our equation is:
Discriminant = (2mb)^2 - 4(1+m^2)(b^2-r^2)
6. Simplify the discriminant:
Discriminant = 4m^2b^2 - 4(b^2-r^2)
= 4m^2b^2 - 4b^2 + 4r^2
= 4[b^2(m^2 - 1) + r^2]
7. Set the discriminant equal to zero and solve for the given condition.
4[b^2(m^2 - 1) + r^2] = 0
8. Dividing both sides of the equation by 4, we get:
b^2(m^2 - 1) + r^2 = 0
9. Rearranging the terms and factoring out b^2, we obtain:
b^2 = -r^2(m^2 - 1)
10. Finally, simplify to reach the desired result:
r^2(1 + m^2) = b^2 [Note: we multiplied both sides by -1]
Therefore, we have shown that r^2(1+m^2) = b^2.
b. To find the coordinates of the point of tangency (-r^2m/b, r^2/b), we use the equation of the tangent line y = mx + b.
1. Substitute x = -r^2m/b into the equation y = mx + b.
y = m(-r^2m/b) + b
= -mr^2m/b + b
= -r^2m^2/b + b
= r^2/b (1 - m^2)
2. The resulting y-coordinate is r^2/b (1 - m^2), so the point of tangency is (-r^2m/b, r^2/b).
c. To prove that the tangent line is perpendicular to the line containing the circle and the point of tangency, we need to show that the product of the slopes of the two lines is -1.
1. The equation of the tangent line is y = mx + b, which has a slope of m.
2. The line containing the circle and the point of tangency has the coordinates (0,0) and (-r^2m/b, r^2/b). The slope of this line can be found using the formula:
Slope = (change in y) / (change in x)
= (r^2/b - 0) / (-r^2m/b - 0)
= r^2 / (-r^2m/b)
= -r^2 / (r^2m/b)
= -b/m
3. Multiply the two slopes: m * (-b/m) = -b/b = -1.
Therefore, we have demonstrated that the tangent line is perpendicular to the line containing the circle and the point of tangency.