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January 1st, 2013, 05:01 AM   #1
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[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)
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January 1st, 2013, 02:47 PM   #2
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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.
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January 5th, 2013, 06:41 AM   #3
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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
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January 10th, 2013, 02:09 AM   #4
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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
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