# math question

Discussion in 'Community Discussion' started by wonga1127, Aug 30, 2007.

1. ### wonga1127 macrumors 6502

Joined:
Mar 16, 2006
Location:
Wishing for a magic bus.
#1
Okay, having trouble with this problem.

Show that the equation of the line tangent to the circle x^2 + y^2 = a^2 at the point (X1 , Y1) (thats x sub 1 y sub 1) on the circle is X1(x) + Y1(y) = a^2

Where do I even start?

2. ### zap2 macrumors 604

Joined:
Mar 8, 2005
Location:
Washington D.C
#2
Oh, my. I should know how to do this...

but I've forgotten, sorry.

Someone here is bound to be able to help

3. ### siurpeeman macrumors 603

Joined:
Dec 2, 2006
Location:
the OC
#3
the slope from the center (0,0) to the point of tangency (x1,y1) is equal to y1/x1. the line from the center to the point of tangency is perpendicular to the tangent line, so the slope of the tangent line is equal to the negative reciprocal, or -x1/y1. using point-slope formula, you get an equation of y - y1 = (-x1/y1)(x - x1).

multiplying and rearranging will lead you to (x1)*x + (y1)*y = (x1)^2 + (y1)^2. looking at the original equation, (x1)^2 + (y1)^2 is equal to a^2, so substituting will get you your equation of (x1)^2 + (y1)^2 = a^2.

you're welcome.

4. ### yg17 macrumors G5

Joined:
Aug 1, 2004
Location:
St. Louis, MO
#4

5. ### wonga1127 thread starter macrumors 6502

Joined:
Mar 16, 2006
Location:
Wishing for a magic bus.
#5
thank you.

And the lolcat made me laugh.

edit: okay i did the point slop formula and everything, but when I simplfied and rearranged I got y1y + x1x = x1^2 + y1^2. x^2 +y^2 = a^2. im stuck!