Equation of circle

[匿名]

2012-03-22 5:24 am

1.In each of the following,write down the coordinates of the point of contact of the circle and the straight line, and find the equation of the circle in the standard form.
a)The circle touches the x-axis and its centre is(-2,5)
2.It is given that the equation of a circle is x^2+y^2+4x+6y=0
a)Tind the coordinates of the centre A.
b)Find the equation of the straight line that passes through the originO and is perpendicular to OA.
3.It is given that the circle x^2+y^2+4x-8y+c=0 passes through the point(2,5)
a)Find the value of c
b)If one end-point of a diameter of the circle is (2,5),find the coordinates of the other end-point of the diameter.

回答 (1)

sioieng

2012-03-22 7:04 am

✔ 最佳答案

1. a)
Point of contact = (-2, 0)

Radius of the circle = 5

Equation of the circle :
(x + 2)² + (y - 5)² = 5²
x² + 4x + 4 + y² - 5y + 25 = 25
x² + y² + 4x - 5y + 4 = 0

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2. a)
Circle: x² + y² + 4x + 6y = 0
Coordinates of centre : A(-4/2, -6/2) = A(-2, -3)

2. b)
Slope of OA = (0 + 3)/(0 + 2) = 3/2
Slope of the required line = -1/(3/2) = -2/3

Equation of the required line :
y - 0 = (-2/3)(x - 0)
y = -2x/3
2x + 3y = 0

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3. a)
Put x = 2 and y = 5 into the equation of the circle.
(2)² + (5)² + 4(2) - 8(5) + c = 0
4 + 25 + 8 - 40 + c = 0
c = 3

3. b)
Circle: x² + y² + 4x - 8y + 3 = 0
Centre = (-4/2, 8/2) = (-2, 4)

Let (a, b) be the other end-point of the diameter.
The centre is the mid-point of (a, b) and (2, 5).
(a + 2)/2 = -2 and (b + 5)/2 = 4
a + 2 = -4 and b + 5 = 8
a = -6 and b = 3

Hence, the other end-point of the diameter = (-6, 3)

參考: sioieng

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