Fun with Loci I

Geometry Level pending

Given the points P ( 0 , 2 ) P(0, -2) and Q ( 7 , 4 ) Q(-7, 4) , find the equation of the moving point A A such that P A Q \angle PAQ is always 90 ° 90 \degree .

x 2 + y 2 + 7 x + 2 y + 8 = 0 x^2+y^2+7x+2y+8=0 x 2 + y 2 7 x 2 y + 8 = 0 x^2+y^2-7x-2y+8=0 x 2 + y 2 + 7 x 2 y + 8 = 0 x^2+y^2+7x-2y+8=0 x 2 + y 2 + 7 x 2 y 8 = 0 x^2+y^2+7x-2y-8=0 x 2 + y 2 + 7 x + 2 y 8 = 0 x^2+y^2+7x+2y-8=0 x 2 + y 2 7 x + 2 y 8 = 0 x^2+y^2-7x+2y-8=0 x 2 + y 2 7 x 2 y 8 = 0 x^2+y^2-7x-2y-8=0

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2 solutions

Steven Chase
Jul 18, 2019

The dot product between a vector from ( x , y ) (x,y) to ( P x , P y ) (P_x,P_y) and a vector from ( x , y ) (x,y) to ( Q x , Q y ) (Q_x,Q_y) should be zero :

( Q x x ) ( P x x ) + ( Q y y ) ( P y y ) = 0 ( 7 x ) ( 0 x ) + ( 4 y ) ( 2 y ) = 0 x 2 + y 2 + 7 x 2 y 8 = 0 (Q_x - x)(P_x-x) + (Q_y-y)(P_y-y) = 0 \\ (-7-x)(0-x) + (4-y)(-2-y) = 0 \\ x^2 + y^2 + 7x -2y - 8 = 0

Easiest solution there is, Steven!

tom engelsman - 1 year, 10 months ago
Chris Lewis
Jul 18, 2019

The locus is a circle with diameter P Q PQ - this is an application of Thales's theorem (or in fact of its converse).

The centre of the circle is the midpoint of P Q PQ (ie ( 7 2 , 1 ) \left(-\frac72,1\right) ); the radius is half the distance P Q PQ (ie 85 2 \frac{\sqrt{85}}{2} ). Its equation is then

( x + 7 2 ) 2 + ( y 1 ) 2 = 85 4 \left(x+\frac72 \right)^2 + (y-1)^2=\frac{85}{4}

which after simplifying becomes

x 2 + y 2 + 7 x 2 y 8 = 0 \boxed{x^2+y^2+7x-2y-8=0}

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