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Consider point $A$ in the angle $xoy$ Draw a circle that passes $A$ and is tangent to lines of angle $xoy$.

My attempt:Because it is tangent to both sides of the angle the center of the circle should be on the bisector of the angle.But I can't find a place for the center of the circle that has the same distance from $A$ too.

Taha Akbari
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2 Answers2

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As noted by you, the center of the circle must lie on the angle bisector of $xoy$. We also note that the center of the circle must be equidistant from the point $A$, and from the line $oy$, and so lies on the parabola with focus $A$ and directrix $oy$.

We thus see that the center of the circle is the intersection of the angle bisector of $xoy$, and the parabola with focus $A$ and directrix $oy$.

The answer to this question shows how to construct the intersection of a line and a parabola with a known focus and directrix.

For concreteness, the process is as follows:

Let $\ell$ be the angle bisector of the angle $xoy$. Reflect $A$ about $\ell$ to obtain the point $A^\prime$. Let $A A^\prime$ intersect $oy$ at $B$. Let $C$ be the intersection of $A A^\prime$ and $\ell$.

Construct semi-circles with diameters $A A^\prime$, and $BC$, and let these intersect at $D$. Construct the circle with center $B$ and radius $BD$. This intersects the line $oy$. Construct a line perpendicular to $oy$ that passes through this point of intersection. This perpendicular line intersects $\ell$ at the center of the required circle.

Dylan
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  • I'm sorry I am not a native englis speaker.What do you mean by word's "parabola","focus" and "directrix" – Taha Akbari Jul 11 '17 at 14:16
  • In this case, it is not important precisely what it means. The procedure can be understood, and proven to work, without ever using the words "parabola", "directrix", or "focus". That said, to answer your question, a parabola is a type of conic section. A typical example is the curve given by the equation $y = x^2$. A parabola is the set of all points that are at the same distance from a given line (called the directrix), and a given point (called the focus). – Dylan Jul 11 '17 at 15:48
  • In the link that you gave how did @Marconius conclude that $|Q_1R_1|^2 = |Q_1F||Q_1F'|$? – Taha Akbari Jul 11 '17 at 16:16
  • $Q_1 R_1$ is tangent to the semi-circle with diameter $F F^\prime$ since $Q_1 R_1$ is perpendicular to $C R_1$, which is a radius of the semi-circle. He then uses the "Power of a Point" theorem to conclude that $| Q_1 R_1|^2 = |Q_1 F||Q_1 F^\prime|$. Alternatively, once can notice tht the triangles $\triangle Q_1 F R_1$ and $\triangle Q_1 R_1 F^\prime$ are similar. – Dylan Jul 11 '17 at 16:21
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Let $P$, on the angle bisector, be the center of the circle to be drawn. Then the distance of $P$ from one of the angle sides is the same as the distance of $P$ from $A$. It follows that $P$ belongs to the parabola having $A$ as focus and one of the angle sides as directrix. The intersections of that parabola with the angle bisector give then the two only possible positions for $P$.

Intelligenti pauca
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