可以幫忙解題一下嗎?感恩?

令 C 點座標為 ( p , q )
因為 C 在圓上, 所以 p^2 + q^2 = 5^2 = 25 ..... (1式)

AB之斜率 = m(AB) = 2/( - 2 - √5 + 5 ) = 2/(3-√5) = (3+√5)/2
m(AC) = q/(p+5)
因為 A, B, C 共線, 所以 m(AC) = m(AB)
q/(p+5) = (3+√5)/2 , 即 q : (p+5) = (3+√5) : 2
故可令 q = (3+√5)k , p+5 = 2k
p = 2k-5

將 p , q 代入 (1式) 得:
(2k-5)^2 + [ (3+√5)k ]^2 = 25
4k^2 - 20k + 25 + ( 14 + 6√5 )k^2 = 25
( 18 + 6√5 )k^2 - 20k = 0
k [ ( 18 + 6√5 )k - 20 ] = 0
k = 0 (不合, 因為推得 p = - 5 , 即 A 點的 x 座標, 但之前已假設 p 是 C 點的 x 座標)
k = 20/( 18 + 6√5 ) = 10/( 9 + 3√5 ) = 10( 9 - 3√5 )/36 = 5( 3 - √5 )/6

AC^2
= (p+5)^2 + q^2
= p^2 + 10p + 25 + q^2
= 25 + 10p + 25 , 由(1式)
= 10p + 50
= 10( 2k - 5 ) + 50
= 20k
= 20*5( 3 - √5 )/6
= 50( 3 - √5 )/3
= 50 - (50/3)√5

AC
= √ [ 50 - (50/3)√5 ]
= √ [ ( 150 - 50√5 ) / 3 ]
=√ [ (25/3)( 6 - 2√5 ) ]
= ( 5 / √3 ) * √( 6 - 2√5 )
= ( 5√3 / 3 ) * ( √5 - 1 )
= (5/3)√15 - (5/3)√3

a + b = 5/3 + 5/3 = 10/3 ..... Ans