# Geometry question 2

How to do the following ?

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### 2 個解答

• 最愛解答

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 ...... 

OB = b :

√[(a - 0)² + (b - 2)²] =b

(a)² + (b - 2)² = b²

a² + b² - 4b+ 4 = b²

a² - 4b + 4 = 0 ...... 

 -  :

-12b + 60 = 0

b = 5

Put into  :

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：

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 ...... 

The circle touches x-axis but not y-axis.

Hence, D not equatl to E ...... 

• 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.

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)