Show that the locus of $P$ as $t$ varies is a circle .

Question

A line through the point $(1,0)$ meets the variable line $y=tx$ at right angle at point $P.$

Find in terms of $t$,the coordinates of $P.$

I’ve found the coordinates of $P$ to be $\displaystyle\Big(\frac{1}{1+t^2},\frac{t}{1+t^2}\Big)$

Show that the locus of $P$ as $t$ varies is a circle and state its centre and radius.

How to show that the locus of $P$ as $t$ varies is a circle?

One way is to see that $x^2+y^2-x=0$ i.e. $(\frac1{1+t^2})^2+(\frac{t}{1+t^2})^2-\frac1{1+t^2}=0.$ — Jan-Magnus Økland, Apr 02 '20 at 10:44

score 1 · Accepted Answer · answered Apr 02 '20 at 12:11

You have the variable coordinates of $P$ : $$x=\frac{1}{t^2+1}$$ $$y=\frac{t}{t^2+1}$$

As Jan-Magnus said, to prove that the locus is a circle, it’s a good idea to square and add $x$ and $y$.

$$x^2+y^2 = \frac{1+t^2}{(1+t^2)^2} = \frac{1}{1+t^2}$$ $$x^2 + y^2 - x=0$$ or $$\left(x-\frac12\right)^2 + y^2 = \left(\frac 12 \right)^2$$

which tells us that the center of the circle is $\left(\frac 12,0 \right)$ and the radius is $\frac 12$.

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