User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

January 1st, 2013, 05:01 AM   #1
Newbie

Joined: Jun 2011

Posts: 17
Thanks: 0

[CIRCLE] Equation of tgt. and distance

Q 1:- Find the equation to the tangent at A to the circle , where the radius through A makes an angle with the x-axis.

My attempt:- Since it is given that the line through A and centre makes an angle with x-axis, so the slope of normal at A is and hence the slope of tangent at A is .Let .Expanding the equation of circle, we get
and tangent to it is .
So,
or

Also from two point form of a straight line, equation of tangent is

or

Equations (1) and (2) are same. So on comparing coefficients and constants, we obtain

Obtaining p and q from these equations looks a tough job. I don't know after that Q 2:- A line is drawn through a fixed point to cut the circle at A and B. Then PA X PB is equal to.
Attempt:- I first thought of the line to pass through the centre(it is one of the cases) and then calculating the point of intersection and then applying the distance formula but , it wasn't a good idea. I only know that PA X PB = PC X PD(refer figure)
[attachment=0:3dc7m5za]Untitled.jpg[/attachment:3dc7m5za]
Attached Images Untitled.jpg (21.1 KB, 149 views) January 1st, 2013, 02:47 PM #2 Senior Member   Joined: Jun 2011 Posts: 298 Thanks: 0 Re: [CIRCLE] Equation of tgt. and distance The center of the circle is at AO and the tangent form, so the slope of the tangent line is The coordinate at A is Now you can find tangent line intersection on the -axis. Suppose the result is Suppose the tangent line is . You can find by substituting for and in the equation. January 5th, 2013, 06:41 AM   #3
Newbie

Joined: Jun 2011

Posts: 17
Thanks: 0

Re: [CIRCLE] Equation of tgt. and distance

Quote:
 Originally Posted by Math Dreamer The center of the circle is at AO and the tangent form, so the slope of the tangent line is The coordinate at A is Now you can find tangent line intersection on the -axis. Suppose the result is Suppose the tangent line is . You can find by substituting for and in the equation.
Thanks for striking me the idea of parametric form. But the slope of tangent at A should be . So A (. By slope point form of a straight line, we get

or First question is now figured out. May i get some help on the second question  January 10th, 2013, 02:09 AM   #4
Newbie

Joined: Jun 2011

Posts: 17
Thanks: 0

Re: [CIRCLE] Equation of tgt. and distance

Just to tell, its seems that i have found the solution.http://en.wikipedia.org/wiki/Circle#Theorems. Point #2--->Tangent-secant theorem.
Quote:
 Tangent-secant theorem If a tangent from an external point D meets the circle at C and a secant from the external point D meets the circle at G and E respectively, then DC2 = DG � DE. (Tangent-secant theorem.)
.
So in my case,

:P Tags circle, distance, equation, tgt ,

,

,

,

tgt in distance

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Andy Fang Algebra 7 November 10th, 2013 04:20 AM prasant38 Algebra 7 June 24th, 2011 05:54 PM hailua Algebra 1 April 28th, 2010 03:12 AM football Algebra 1 April 8th, 2010 04:46 AM luftikus143 Algebra 1 February 11th, 2008 06:26 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      