The line 2y-x=12 intersects the circle x2+y2-10x-12y+36=0 at the point A and B. The perpendicular bisector of AB intersects the circle at the point P and Q. What is the exact coordinates of P and Q?

Question

The line 2y-x=12 intersects the circle x2+y2-10x-12y+36=0 at the point A and B. The perpendicular bisector of AB intersects the circle at the point P and Q. What is the exact coordinates of P and Q?

Answer

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Answer 1

In a nutshell

The circle's equation ( x - 5 )² + ( y + 6 ) ² = 25 allows us to locate its center as Ω ( 5, -6 ).

Notice that we not have to find the coordinates of the points A and B since the perpendicular bisector of AB

will pass through the center Ω ( 5, -6 )..

The equation of the perpendicular bisector of AB is y +6 = -2 ( x -5 ) [ Eq 1.]

Substituting back to the equation of the circle -2 ( x -5 ) for y +6 we get

( x -5 )² = 5 ⇒ x = 5 ± √5

Then if x _P = 5 + √5 and from [ Eq 1.] we get y _P = - 2√5 - 6 Hence ( x _P , y _P ) = ( 5 + √5 , - 2√5 - 6 )

In a similar manner find the coordinates of the point Q.