唔知点prove the equation of C?

2015-10-24 7:00 am

The coordinates of the points P and Q are(4,-1)and(-14,23)
a) let L be the perpendicular bisector of PQ.
(i)Suppose that G is a point lying on L. Denote the x-coordinate of G by h.Let C be the circle which is centered at G and pass through P and Q. Prove that the equation o C is 2x^2 +2y^2-4hx-(3h+59)y+13h-93=0

回答 (1)

Lopez

2015-10-24 8:08 am

✔ 最佳答案

pf :
Since L ⊥ PQ , we have
m(L) * m(P,Q) = - 1
m(L) * (23+1)/(-14-4) = - 1
m(L) * 24/(-18) = - 1
m(L) = 18/24 = 3/4

Let M be the mid-point of P and Q ,
then M on L , and
M = ( (4-14)/2 , (-1+23)/2 ) = ( - 5 , 11 )
Thus, the eq. of L is:
y-11 = (3/4)(x+5)

Let x = h,
y = 11 + (3/4)h + 15/4 = (3/4)h + 59/4
Let u = (3/4)h + 59/4
G = ( h , u ) = ( h , (3/4)h + 59/4 )

Let R be the radius of circle C , then
R^2 = d(G,P)^2 = (h-4)^2 + (u+1)^2

Hence, eq. of circle C is:
(x-h)^2 + (y-u)^2 = R^2 = (h-4)^2 + (u+1)^2
x^2 - 2hx + h^2 + y^2 - 2uy + u^2 = h^2 - 8h + 16 + u^2 + 2u + 1
x^2 + y^2 - 2hx - 2uy + 8h - 2u - 17 = 0
x^2 + y^2 - 2hx - 2[ (3/4)h + 59/4 ]y + 8h - 2[ (3/4)h + 59/4 ] - 17 = 0
2x^2 + 2y^2 - 4hx - (3h+59)y + 16h - 3h - 59 - 34 = 0
2x^2 + 2y^2 - 4hx - (3h+59)y + 13h - 93 = 0

Q.E.D.

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