F5 Equations of circles

Anonymous

2011-02-14 2:51 am

A circle passes through (5, -3) and (-6, 2), and
its centre lies on the line 3x - y +5 = 0. Find its equation. Give the answer in standard form ( centre-radius form).

更新1:

For The slope of OM = (t + 1/2)/(s + 1/2) Where is M ?

回答 (1)

myisland8132

2011-02-14 3:09 am

✔ 最佳答案

mid-point of A(5,-3) and B(-6,2) is (-1/2,-1/2)

Let the centre O(s,t)

The slope of AB = -5/11

The slope of OM = (t + 1/2)/(s + 1/2)

Since m_AB*m_OM = -1

(-5/11)[(t + 1/2)/(s + 1/2)] = -1

5(2t + 1) = 11(2s + 1)

Also, from 3s - t + 5 = 0 => t = 3s + 5

So, 5(6s + 10 + 1) = 11(2s + 1)

30s + 55 = 22s + 11

-8s = 44

s = -5.5 and then t = -11.5 M(-5.5,-11.5)

AM = √[(-5.5 - 5)^2 + (-11.5 + 3)^2] = √182.5

So, the equation of the circle is :

(x + 5.5 )^2 + (y + 11.5)^2 = 182.5

2011-02-16 13:06:34 補充：
M = mid - point of AB

👍 0 👎 0