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Power of a point with respect to a circle

Posted November 29th, 2008 by Structure

Two lines by a point P outside a circle cut in A,B respectively in C and D. Then

$PA*PB=PC*PD=TP^2$

$\triangle PAC\sim\triangle PDB$

as $\angle PBD=\angle PCA\:\:;\angle PDB=\angle PAC$

$\frac{PA}{PD}=\frac{PC}{PB}$

Posted October 19th, 2008 by Structure

We solve the differential equation

$y"(t)+y(t)=f(t)$

with initial condition

$y(0)=0,\:y'(0)=0$

where

$f(t)=u(t-2\pi)-u(t-\pi)$

$\mathcal{L}(y(t)(s)=Y(s)$

$\mathcal{L}(y'(t)(s)=sY(s)$

$\mathcal{L}(y"\;(t)(s)=s^2Y(s)$

We have

$\mathcal{L}(y"(t)+y(t))(s)=s^2Y(s)+Y(s)=F(s)=\frac{1}{s}(e^{-2\pi s}-e^{-\pi s})$

We have

$Y(s)=\frac{1}{s^2+1}F(s)$

We have from convolution theorem

$y(t)=\sin*f(t)=\int_{0}^{t}f(\tau)\sin(t-\tau)d\tau$

Posted October 5th, 2008 by Structure