# F.5 Maths equation of circle

Show that the 2 circles C1: x^2 +y^2 -16x-12y+75=0 and

C2: x^2 +y^2 +2x+12y-63=0 touch each other.

My calculation:

http://postimg.org/image/6rxfgpbfx/

I have some questions about my explanation(the red rectangle part).

I think that it is not detail enough. Also, the explanation only suitable for the case

that the circles touh externally. When they touch internally, the explanation is

wrong. Therefore, I want to ask for a better solution for this question, thank you!!

### 1 個解答

• 最愛解答

Let O1 and O2 be the centres of the C1 and C2 respectively.

C1 : x² + y² - 16x - 12y + 75 = 0

Coordinates of O1 = (16/2, 12/2) = (8, 6)

Radius of C1 = √[(-16/2)² + (-12/2)² - 75] = 5

C2 : x² + y² + 2x + 12y - 63= 0

Coordinates of O 2 = (-2/2, -12/2) = (-1, -6)

Radius of C2 = √[(2/2)² + (12/2)² - (-63)]= 10

O1O2 = √[(8 + 1)² +(6 + 6)²] = 15

The distance between the two centres isequal to the sum of the two radii.

Hence, the two circles touchexternally.

2015-03-18 20:06:13 補充：

Alternative method :

C1 : x² + y² - 16x - 12y + 75 = 0 ...... 

C2 : x² + y² + 2x + 12y - 63 = 0 ...... 

 -  :

18x + 24y - 138 = 0

3x + 4y - 23 = 0

x = (23 - 4y)/3 ...... 

2015-03-18 20:06:40 補充：

Put  into  :

[(23 - 4y)/3]² + y² + 2[(23 - 4y)/3] + 12y - 63 = 0

(23 - 4y)² + 9y² + 6(23 - 4y) + 108y - 567 = 0

529 - 184y + 16y² + 9y² + 138 - 24y + 108y - 567 = 0

25y² - 100y + 100 = 0

y² - 4y + 4 = 0

(y - 2)² = 0

y = 2 (double root)

2015-03-18 20:06:59 補充：

The two circles meet at only one point.

Hence, the two circles touch each other.