A-maths!

Let the equation of the required circle be:
x² + y² - 6y + k = 0

(Since the required circle has the same centre as the circle x² + y² - 6y - 8 = 0)


{x² + y² - 6y + k = 0 ───── (1)
{2x - y + 7 = 0, i.e. y = 2x + 7 ───── (2)

Sub (2) into (1):

x² + (2x + 7)² - 6(2x + 7) + k = 0
5x² + 16x + (7 + k) = 0

Since they touch each other at one point,

∴ Δ = 0

i.e. (16)² - 4(5)(7 + k) = 0
256 - 140 - 20k = 0
k = 29/5

So, the equation of the required circle:

x² + y² - 6y + 29/5 = 0

5x² + 5y² - 30y + 29 = 0

圓心 = (0,3)
半徑 = ! 2(0) - 3 + 7 ! / sqrt(5) = 4 / sqrt(5)

方程
x^2 + (y - 3)^2 = 16/5
化簡, 得答案