equations of circles 急要

1.
For the circle (x - 3)² + (y + 2)² = 25, the centre is (3, -2)

Let B = (h, k), and the centre is the mid-point of AB.
(h + 7)/2 = 3 and (k + 1)/2 = -2
h + 7 = 6 and k + 1 = -5
x = -1 and k = -5

Hence, B = (-1, -5)


= = = = =
2.
Suppose that G is the centre of the circle. Otherwise, please inform.

Centre of the circle = (5, -2)
Slope of BG = (-1 + 2)/(3 - 5) = -1/2
Slope of the required line = -1/(-1/2) = 2

The required line:
(y + 1)/(x - 3) = 2
2x - 6 = y + 1
2x - y - 7 = 0


= = = = =
3.(a)
For the cicle x² + y² + 12x - 16y = 0
Let O = centre
O = (-6, 8) and radius = √[(12/2)² + (16/2)²] = √100 = 10

OA = √[(3 + 6)² + (7 - 8)²] = √82 < radius, thus A is inside the circle.
OB = √[(-11+6)² + (0 - 8)²] = √89 < radius, thus B is inside the circle.
Hence, AB is inside the circle.

3.(b)
Slope of AB = (7 - 0)/(3 + 11) = 1/2
Slope of the required line = 1/2

The required line:
(y - 8)/(x + 6) = 1/2
x + 6 = 2y - 16
x - 2y + 22 = 0


= = = = =
4.(a)
(4x - 1)²+ 16(y + 3)² - 80 = 0
[(4x - 1)²+ 16(y + 3)² - 80]/16 = 0
[x - (1/4)]² + (y + 3)² = (√5)²
The equation is in the form of (x - h)² + (y - k)² = r²
It is a real circle.

4.(b)
Centre = (-(-1/4), -3) = (1/4, -3)


= = = = =
5.
L : x + y - 6 = 0 ... [1]
C: x² + y² - 18 = 0 ... [2]

From [1]:
x = 6 - y ... [3]

Put [3] into [1]:
(6 - y)² + y² - 18 = 0
36 - 12y + y² + y² - 18 = 0
2y² - 12y + 18 = 0
y² - 6y + 9 = 0

Determinant Δ = (-6)² - 4*1*9 = 0
L and C has only one point of intersection.
Hence, L and is a tangent to C.

火星的 Q.3b 也代錯了公式，變成了：
(y - xo)/(x - yo) = slope

In Q.1 of 火星......
When (1/2)(y + 1) = -2
y = 4 ?

Q1
from the equation of the circle (x-3)^2 + (y+2)^2 = 25, we know that the centre of the circle is (3,-2).
Let B=(x,y).
∵ The centre of the cirlce is the mid-point of AB,
∴ ½ (x+7) = 3 and ½ (y+1) = -2
Solve the above equations, we have: x = -1 and y = 4
∴ The coordinates of point B is (-1,4).

Q2
What is G?
Is it a point on the given circle??
Also, the question has (a) but not (b), (c), ... ??

Q3
(a)
In another words, the question asks whether both A and B are inside the given circle. The given equation: x^2+y^2+12x-16y=0
i.e. (x+6)^2+(y-8)^2=100
shows that the centre of the circle is (-6,8) and the radius is 10.
Distance between A and the centre of the circle
= { [3-(-6)]^2+(7-8)^2 }^½
= √82
<10
Distance between B and the centre of the circle
= { [-11-(-6)]^2+(0-8)^2 }^½
= √89
<10
∵ Both points A and B lie inside the given circle,
∴ The line segment joining A(3,7) and B(-11,0) is inside the circle.
(b)
The slope of the line AB = (0-7)/(-11-3) = 7/8
∴ The equation of the straight line that passes through the centre of the circle and is parallel to AB is: (y-(-6))/(x-8) = 7/8
8(y+6) = 7(x-8)
8y+48 = 7x-56
7x - 8y - 104 = 0

Q4
(a) The equation (4x-1)^2 + 16(y+3)^2 - 80 = 0
(equivalent) 16(x-¼)^2 + 16(y+3)^2 - 80 = 0
(equivalent) (x-¼)^2 + (y+3)^2 - 5 = 0
(equivalent) (x-¼)^2 + (y+3)^2 = 5
represents a real circle, with centre (¼,-3) and radius √5
(b)
As mentioned, the coordinates of the centre of the circle is (¼,-3).

Q5
The straight line L : x+y-6=0 is a tangent to the circle C : x^2+y^2-18=0
if and only if they have exacly one intersection, i.e. they touch each other.
L: x+y-6=0 becomes y=6-x
Sub. y=6-x into C: x^2+y^2-18=0
x^2+(6-x)^2-18=0
x^2+36-12x+x^2-18=0
2(x-3)^2=0
The system of equations has exactly one root, i.e. the intersection of L and C is (3,3).
∴ L is tangent to C.