A.Maths - 圓

Since the circle touches both axes, if we let the radius be r, the coordiates of the center is (r,r). So the equation is (x-r)^2 + (y-r)^2 = r^2

Substitute x=8 and y=1 into the equation,
　　　(8 - r)^2 + (1 - r)^2 = r^2
　　　64 - 16r + 1 - 2r + r^2 = 0
　　　r^2 - 18r + 65 = 0
　　　(r - 5)(r - 13) = 0
　　　r = 5 or r = 13
So the equation required is
(x - 5)^2 + (y - 5)^2 = 25 or (x - 13)^2 + (y - 13)^2 = 169

2007-07-14 17:32:27 補充：
According to my A.Maths teacher, answers in the form (x - h)^2 (y - k)^2 = r^2 (standard form) or x^2 y^2 Dx Ey F = 0 (general form) are both acceptable.If you like, you may expand my answer to;x^2 y^2 -10x - 10y 25 = 0 orx^2 y^2 -26x - 26y 169 = 0 or

2007-07-14 17:33:49 補充：
Something wrong ...The "plus" sign cannot be shown ...Please note that ...

Let the equation of the circle be:

x² + y² + Dx + Ey + F = 0



Since the circle passes through (8 , 1).

So, (8)² + (1)² + D(8) + E(1) + F = 0

8D + E + F = -65 ─── (1)


Since the circle touches the x-axis

{ x² + y² + Dx + Ey + F = 0 ─── (2)
{ y = 0 ─── (3)

Sub (3) into (2):

∴ x² + Dx + F = 0

Δ = 0

i.e. D² - 4(1)F = 0

D² - 4F = 0 ─── (4)


Since the circle touches the y-axis

{ x² + y² + Dx + Ey + F = 0 ─── (2)
{ x = 0 ─── (5)

Sub (5) into (2):

∴ y² + Ey + F = 0

Δ = 0

i.e. E² - 4(1)F = 0

E² - 4F = 0 ─── (6)



So, D = ±2√F

and E = ±2√F


Since the circle passes through (8 , 1) and touches the x-axis and y-axis.

So, the centre of the circle lies in the first quadrant.


So, D = E = 2√F

8D + E + F = -65 ─── (1)

8(2√F) + (2√F) + F = -65

F + 18√F + 65 = 0

√F = -5 or -13

F = 25 or 169

When F = 25, then D = E = -10

When F = 169, then D = E = -26


So, the equations of the circle are:

x² + y² - 10x - 10y + 25 = 0 or x² + y² -26x - 26y + 169 = 0