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A(-7,0) and B(7,0) are two fixed points. Find the locus of a moving point
P at which AB subtend a right angle.

​

Sagot :

jacob193

Answer:

[tex]\angle {\rm APB} = 90^{\circ}[/tex] as long as point [tex]P[/tex] is on the circle of radius [tex]r = 7[/tex] centered at [tex](0,\, 0)[/tex] (i.e., the position of point [tex]P[/tex] [tex](x,\, y)[/tex] should satisfy [tex]x^{2} + y^{2} = 7^{2}[/tex].)

Step-by-step explanation:

By Thale's Theorem, if a point [tex]{\rm P}[/tex] is on a circle, endpoints [tex]{\rm A}[/tex] and [tex]{\rm B}[/tex] of any diameter [tex]{\rm AB}[/tex] of that circle would subtend a right angle with point [tex]{\rm P}[/tex], provided that [tex]{\rm P}[/tex] isn't on that diameter. In other words, as long as [tex]{\rm P}[/tex] isn't on the diameter [tex]{\rm AB}[/tex] of that circle, [tex]\angle {\rm APB} = 90^{\circ}[/tex].

The converse of this theorem is also true: if the two endpoints [tex]{\rm A}[/tex] and [tex]{\rm B}[/tex] of a line segment [tex]{\rm AB}[/tex] subtends a right angle with a third point [tex]{\rm P}[/tex] (i.e., [tex]\angle {\rm APB} = 90^{\circ}[/tex],) then point [tex]{\rm P}[/tex] must be on the (unique) circle where [tex]{\rm AB}[/tex] is a diameter.

This question is asking for the possible positions of point [tex]{\rm P}[/tex] such that the angle [tex]\angle {\rm APB}[/tex] would be a right angle ([tex]90^{\circ}[/tex].) Observe that this requirement of this question resembles the premise of the converse of Thale's Theorem. Hence, by the converse of Thale's Theorem, point [tex]{\rm P}[/tex] must be on the circle where segment [tex]{\rm AB}[/tex] is a diameter.

The center of the circle where segment [tex]{\rm AB}[/tex] is a diameter is the same as the center of segment [tex]{\rm AB}[/tex]. Given that [tex]{\rm A}[/tex] is at [tex](-7,\, 0)[/tex] while [tex]{\rm B}[/tex] is at [tex](7,\, 0)[/tex], the center of that circle would be [tex](0,\, 0)[/tex]. The radius of the circle is one-half the length of its diameter: [tex]r = 7[/tex].

Hence, the equation for this circle would be:

[tex](x - 0)^{2} + (y - 0)^{2} = 7^{2}[/tex].

Simplify to obtain:

[tex]x^{2} + y^{2} = 7^{2}[/tex].

In other words, [tex]{\rm P}[/tex] is on the circle [tex]x^{2} + y^{2} = 7^{2}[/tex] (where [tex]{\rm AB}[/tex] is a diameter) if and only if [tex]\angle {\rm APB} = 90^{\circ}[/tex]:

  • [tex]\text{P is on the circle} \implies \angle {\rm APB} = 90^{\circ}[/tex] is from Thale's Theorem.
  • [tex]\angle {\rm APB} = 90^{\circ} \implies \text{P is on the circle}[/tex] is from the converse of Thale's Theorem.
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