✔ 最佳答案
1. C: (x - 2)^2 + y^2 = 5 ... (1)
L: x + 2y = 7 ... (2)
From equation (2), we have x = 7 - 2y ... (3)
Put (3) into (1):
[(7 - 2y) - 2]^2 + y^2 = 5
(5 - 2y)^2 + y^2 = 5
25 - 20y + 4y^2 + y^2 = 5
y^2 - 4y + 4 = 0 ... (*)
Dicriminant of (*) = (-4)^2 - 4(1)(4) = 0
Hence, the line x + 2y = 7 and the circle (x - 2)^2 + y^2 = 5 touches at only one point.
So, x + 2y = 7 is a tangent to (x - 2)^2 + y^2 = 5.
2. Let x^2 + y^2 + Dx + Ey + F = 0 be the equation of the circle.
We consequently put A(2 , 8), B(-3 , 3) and C(6 , 6) into the equation:
(2)^2 + (8)^2 + 2D + 8E + F = 0
2D + 8E + F = -68 ... (1)
(-3)^2 + (3)^2 - 3D + 3E + F = 0
-3D + 3E + F = -18 ... (2)
(6)^2 + (6)^2 + 6D + 6E + F = 0
6D + 6E + F = -72 ... (3)
Solving equations (1), (2) and (3), we have D = -4, E = -6, F = -12
So, the equation of the circle is x^2 + y^2 - 4x - 6y - 12 = 0.
3. If AC is the diameter of the circle, then the centre of the circle would be the mid-point of the line AC.
Let C(x , y)
1 = (-2 + x) / 2, x = 4
2 = (-2 + y) / 2, y = 6
So, C(4 , 6).
By noting the diagram of the circle, we see angle ABC is a right-angle (angle of semi-circle)
Hence, triangle ABC is a right-angles triangle.
AB = √[(-2 - 5)^2 + (-2 - (-1))^2] = 5√2
BC = √[(5 - 4)^2 + (-1 - 6)^2] = 5√2
Hence, area of triangle ABC
= 1/2 X AB X BC
= 1/2 (5√2)^2
= 25 sq.units.
