Find the images of the circles $(x-2)^2+y^2=4$ and $(x-2)^2+y^2=9$ under reciprocation with respect to the unit circle.
This is my final exam question but I cannot do any thing about it. Please help me.
Reciprocation with respect to the unit circle
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geometry
projective-geometry
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1Is this a currently on-going exam, or was it the last problem on an exam that is finished, and you just want to know how to do it? – 2017-01-10
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0Reciprocation is a Moebius map: this sends circles into circles (where straight lines are considered circles of radius $\infty$). So, in particular, you should have been looking for some circles in the plane. – 2017-01-10
1 Answers
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Take for example any point $\;(a,b)\;$ in the first circle: $\;(a-2)^2+b^2=4\;$ . We want a point $\;(p,q)\;$ on the ray $\;(2,0)\to(a,b)\;$ , meaning: on the line
$$(2,0)+t(a,b)=(at+2,bt)\;,\;\;0\le t\in\Bbb R$$
s.t. $\;d((a,b),(2,0))\cdot d((p,q),(2,0))=1^2=1\;$ (with $\;d=\;$ distance) , and this means $\; d((p,q),(2,0))=\cfrac12\;$ .
Now, write $\;(p,q)=(at+2,bt)\;$ , for some $\;0\le t\in\Bbb R\;$ ,so we get
$$\frac14=d((p,q),(2,0))^2=a^2t^2+b^2t^2\implies t=\frac1{2\sqrt{a^2+b^2}}$$
and your point is
$$\left(\,\frac a{2\sqrt{a^2+b^2}}+2\,,\,\,\frac b{2\sqrt{a^2+b^2}}\,\right)$$
Understand the above and fill in details, and do the other case.