Equation of circle

2008-06-29 7:04 pm

A familt of circles is given by the equation x^2 +y^2 -4x +3 +k(x-y-1) = 0
where k is a any constant.

a. Find the equation of the circle in the family which passes through
the point(2,0)
b. Find the equation of the circle in the family with radius 開方13

回答 (1)

Audrey Hepburn

2008-06-29 7:34 pm

✔ 最佳答案

Equation of family of circles:

x2 + y2 - 4x + 3 + k(x - y - 1) = 0

a. For the circle passes through (2 , 0)

(2)2 + 0 - 4(2) + 3 + k(2 - 0 - 1) = 0

k = 1

So, the required equation of the circle:

x2 + y2 - 4x + 3 + (x - y - 1) = 0

x2 + y2 - 3x - y + 2 = 0

b. Rewrite the equation of the family of circles:

x2 + y2 + (k - 4)x - ky + (3 - k) = 0

Radius = √13

i.e. 1/2 √[(k - 4)2 + (-k)2 - 4(3 - k)] = √13

(k2 - 8k + 16) + k2 - 12 + 4k = 4(13)

k2 - 2k - 24 = 0

(k - 6)(k + 4) = 0

k = -4 or 6

Therefore, the equation of circles:

x2 + y2 + (6 - 4)x - 6y + (3 - 6) = 0 or x2 + y2 + (-4 - 4)x - (-4y) + [3 - (-4)] = 0

x2 + y2 + 2x - 6y - 3 = 0 or
x2 + y2 - 8x + 4y + 7 = 0

參考: Myself~~~

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