My Math Forum [CIRCLE] Equation of tgt. and distance
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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 $(x-a)^2+(y-b)^2=r^2$, where the radius through A makes an angle $\alpha$ 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 $\tan\alpha$ and hence the slope of tangent at A is $\frac{-1}{\tan\alpha }=-\cot\alpha$.Let $A(p,q)$.Expanding the equation of circle, we get
$x^2+y^2-2ax-2by+(a^2+b^2-r^2)$ and tangent to it is $T=0$.
So, $xp+yq-a(x+p)-b(y+q)+(a^2+b^2-r^2)$
or$x(p-a)+y(q-b)+(a^2+b^2-r^2-ap-bq)-------(1)$

Also from two point form of a straight line, equation of tangent is
$y-q=-\cot \alpha (x-p)$
or $x\cot \alpha +y-(q+p\cot \alpha )-------(2)$

Equations (1) and (2) are same. So on comparing coefficients and constants, we obtain
$\frac{p-a}{\cot \alpha}=\frac{q-b}{1}=\frac{a^2+b^2-r^2-ap-bq}{-(q+p\cot \alpha )}$

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 $P(\alpha ,\beta )$ to cut the circle $x^2+y^2=r^2$ 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]
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 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 $O(a,b)$ AO and the tangent form$\frac{\pi}{2}$, so the slope of the tangent line is $m=(\frac{\pi}{2}-\alpha )$ The coordinate at A is $(\ a-r\ \cos \alpha, \ b+r\ \tan \alpha)$ Now you can find tangent line intersection on the $x$-axis. Suppose the result is $(x_1, 0)$ Suppose the tangent line is $y=mx+C$. You can find $C$ by substituting $(x_1, 0)$ for $x$ and $y$ in the equation.
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 $O(a,b)$ AO and the tangent form$\frac{\pi}{2}$, so the slope of the tangent line is $m=(\frac{\pi}{2}-\alpha )$ The coordinate at A is $(\ a-r\ \cos \alpha, \ b+r\ \tan \alpha)$ Now you can find tangent line intersection on the $x$-axis. Suppose the result is $(x_1, 0)$ Suppose the tangent line is $y=mx+C$. You can find $C$ by substituting $(x_1, 0)$ for $x$ and $y$ in the equation.
Thanks for striking me the idea of parametric form. But the slope of tangent at A should be $-\cot\alpha$. So A ($a\pm r\cos\alpha, b \pm r\sin\alpha$. By slope point form of a straight line, we get
$y-(b \pm r\sin\alpha)=-\cot\alpha (x-(a \pm r\sin\alpha))$
or$(x-a)\cos\alpha+(y-b)\sin\alpha=\pm r$
First question is now figured out. May i get some help on the second question

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,
$PA X PB= (Length\; of\; tangent\; from\; P)^2=(\sqrt(S_1))^2=\alpha ^2+\beta ^2-r^2$
:P

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