Geometry question 2

3.
Put y = 0 (x-axis) into the equation (x + 3)² +(y - 4)² = 25
(x + 3)² + (0 - 4)² =25
(x + 3)² + 16 = 25
(x + 3)² = 9
x + 3 = 3 or x + 3 = -3
x = 0 and x = -6
Hence, OA = 0 - (-6) = 6

Put x = 0 (y-axis) into the equation (x + 3)² +(y - 4)² = 25
(0 + 3)² + (y - 4)² =25
9 + (y - 4)² = 25
(y - 4)² = 16
y - 4 = 4 or y - 4 = -4
y = 8 or y = 0
Hence, OC = 8 - 0 = 8

Area of rectangle OABC
= OA * OC
= 6 * 8
= 48 ...... answer : B


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4.
Let O(a, b) be the centre of the circle.
The circle touches the positive x-axis at C(a, 0).
Radius of the circle = b

OA = b :
√[(a - 0)² + (b - 8)²] =b
(a)² + (b - 8)² = b²
a² + b² -16b + 64 = b²
a² - 16b + 64 = 0 ...... [1]

OB = b :
√[(a - 0)² + (b - 2)²] =b
(a)² + (b - 2)² = b²
a² + b² - 4b+ 4 = b²
a² - 4b + 4 = 0 ...... [2]

[1] - [2] :
-12b + 60 = 0
b = 5

Put into [1] :
a² - 16(5) + 64 = 0
a² = 16
a = 4 or a = -4 (rejected)

Centre = (4, 5) and radius = 5
The equation :
(x - 4)² + (y - 5)² =5²
x² - 8x + 16 + y² -10y + 25 = 25
x² + y² -8x - 10y + 16 = 0 ...... answer : (A)

2015-03-04 21:59:46 補充：
4. Alternative answer :
Let x² + y² + Dx + Ey + F = 0 be the required equation.

Centre (-D/2, -E/2) is in the first quadrant.
-D/2 > 0 and -E/2 > 0
Hence, D < 0 and E < 0 ...... [1]

The circle touches x-axis but not y-axis.
Hence, D not equatl to E ...... [2]

The answer : A

3.
Put x = 0,
(y - 4)² = 25 - 9 = 16
y - 4 = 4 or -4
y = 0 or 8
OC = 8

Put y = 0,
(x + 3)² + 16 = 25
(x + 3)² = 9
x + 3 = 3 or -3
x = 0 or -6
OA = 6

Area = 8 * 6 = 48

(B)

2015-03-04 21:51:09 補充：
4.
Let the centre be C = (a, b)
By symmetry,
b = (2 + 8)/2 = 5

Then, we note that radius = 5 also.
This is because the circle touches x-axis.

Denote the mid-point of AB be M.

Consider right-angled triangle AMC.
3² + a² = (radius)²
9 + a² = 25
a = 4

2015-03-04 21:51:28 補充：
With centre (4, 5) and radius 5, the equation is
(x - 4)² + (y - 5)² = 5²
x² + y² - 8x - 10y + 16 = 0

(A)