Reply
Fri 12 Aug, 2005 04:08 pm
The tangent line 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 if the circle is
x^2 + y^2 = r^2 and the equation of the tangent line is y =
mx + b,
a) show that r^2(1 + m^2) = b^2
b) the point of tangency is (-r^2m/b, r^2/b)
c) the tangent line is perpendicular to the line containing
the center of the circle and the point of tangency.
HINT: The quadratic equation x^2 + (mx + b) ^2 = r^2 has
exactly one solution.
ALSO:
There is a picture of a circle with radius r going from the
center (0,0) to the point of tangency on the xy-plane
I worked on this question for over 60 minutes but to no avail.
Hmm I've had a quick bash at this and don't find an obvious solution but i haven't done circle theorems in about a year so I'm a bit rusty
I have a question though for anybody else watching the thread:
What is with part c? It's one of the primary circle theorems that a tangent will always meet the circle's radius at a 90 degree angle - ie they will always be perpendicular. How can the question ask you to prove one of the rules you're using to solve the problem?!
Also just had a quick think and realised (slightly obviously possibly but i'm tired so not sure) this:
y^2 = r^2 - x^2
(mx + b)^2 = r^2 - x^2
This equation will have a route where the squares of (mx + b) and x together = the square of r (because the root occurs where it all equals 0)
... pants... that was leading somewhere and then i got a sign wrong and aaargh
a) show that r^2(1 + m^2) = b^2
HINT: The quadratic equation x^2 + (mx + b) ^2 = r^2 has
exactly one solution.
The hint is true because the tangent intersects the circle at exactly one point. Since there is exactly one solution, the discriminant (B^2 - 4AC) must be zero.
x^2 + (mx+b)^2 - r^2 = 0
x^2 + m^2x^2 + 2bmx + b^2 - r^2 = 0
(1+m^2)x^2 + 2bmx + (b^2 - r^2) = 0
The discriminant is
4b^2m^2 - 4(1+m^2)(b^2 - r^2)
Set it to zero:
4b^2m^2 - 4(1+m^2)(b^2 - r^2) = 0
b^2m^2 - (1+m^2)b^2 + (1+m^2)r^2 = 0
(1+m^2)r^2 = (1+m^2)b^2 - b^2m^2
(1+m^2)r^2 = b^2
Bekaboo wrote:I have a question though for anybody else watching the thread:
What is with part c? It's one of the primary circle theorems that a tangent will always meet the circle's radius at a 90 degree angle - ie they will always be perpendicular. How can the question ask you to prove one of the rules you're using to solve the problem?!
Once you have the tangent, you can calculate the slope of the radius through the tangent. Ignoring vertical and horizontal lines, two lines are perpendicular if the product of their slopes is -1.
b) the point of tangency is (-r^2m/b, r^2/b)
Given Ax^2 + Bx + C = 0,
the quadratic formula tells us that
x = [-B +/- sqrt(B^2 - 4AC)] / 2A
However, as mentioned before, the discriminant is zero.
Therefore,
x = -B/2A
In this case, that's
x = -2bm/2(1+m^2) = -bm/(1+m^2) = -b^2m/b(1+m^2)
In (a) we showed that r^2(1+m^2) = b^2
So, we have
x = -r^2(1+m^2)m/b(1+m^2) = -r^2m/b
Since y = mx+b
y = -r^2m^2/b + b = -r^2m^2/b + b^2/b = (b^2 - r^2m^2)/b
Since b^2 = r^2(1+m^2) = r^2 + r^2m^2
we have
y = (r^2 + r^2m^2 - r^2m^2)/b = r^2/b
hey
I want to thank all those who replied, especially markr.
Your notes are great for study time.
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace