Implicit differentiation ?

(x - 6)² + (y - 17)² = 100

x² + y² = 225


Point of intersection

(x - 6)² + (y - 17)² = 100

x² - 12x + 36 + y² - 34y + 289 = 100

x² + y² - 12x - 34y = - 225 → recall the other circle: x² + y² = 225

225 - 12x - 34y = - 225

- 12x - 34y = - 450

6x + 17y = 225 ← equation of the line that contents the points of intersection between the 2 circles


6x + 17y = 225

6x = 225 - 17y

x = (225 - 17y)/6

x² = (225 - 17y)²/6²

x² = (50625 - 7650y + 289y²)/36 → recall the other circle: x² + y² = 225 → x² = 225 - y²

225 - y² = (50625 - 7650y + 289y²)/36

36.(225 - y²) = 50625 - 7650y + 289y²

8100 - 36y² = 50625 - 7650y + 289y²

325y² - 7650y = - 42525

13y² - 306y = - 1701

y² - (306/13).y = - 1701/13

y² - (306/13).y + (153/13)² = - (1701/13) + (153/13)²

y² - (306/13).y + (153/13)² = 36²/13²

[y - (153/13)]² = (36/13)²

y - (153/13) = ± 36/13

y = (153/13) ± (36/13)

y = (153 ± 36)/13


y₁ = (153 + 36)/13 = 189/13 → recall the line: 6x + 17y = 225

6x₁ = 225 - 17y₁

6x₁ = 225 - 17.(189/13)

6x₁ = - 288/13

x₁ = - 48/13

→ Point (- 48/13 ; 189/13)


y₂ = (153 - 36)/13 = 9 → recall the line: 6x + 17y = 225

6x₂ = 225 - 17y₂

6x₂ = 225 - 153

6x₂ = 72

x₂ = 12

→ Point (12 ; 9) ← this is the given point


Recall the first circle:

(x - 6)² + (y - 17)² = 100

(y - 17)² = 100 - (x - 6)²

(y - 17)² = 100 - (x² - 12x + 36)

(y - 17)² = 100 - x² + 12x - 36

(y - 17)² = - x² + 12x + 64

y - 17 = ± (- x² + 12x + 64)^(1/2)

y = ± (- x² + 12x + 64)^(1/2) + 17

y' = ± (1/2).(- 2x + 12) * (- x² + 12x + 64)^[(1/2) - 1]

y' = ± (- x + 6) * (- x² + 12x + 64)^(- 1/2)

y' = ± (x - 6) / (- x² + 12x + 64)^(1/2)

y' = ± (x - 6) / √(- x² + 12x + 64) ← this is the slope of the tangent line (ℓ₁) to the circle at x

y' = ± (x - 6) / √(- x² + 12x + 64) → when: x = 12

y' = ± (12 - 6) / √(- 144 + 144 + 64)

y' = ± 6/8

y' = ± 3/4 → you can see on the drawing, that the slope is positive

y' = 3/4


Recall the first circle:

x² + y² = 225

y² = 225 - x²

y = ± (225 - x²)^(1/2)

y' = ± (1/2) * (- 2x) * (225 - x²)^[(1/2) - 1]

y' = ± (- x) * (225 - x²)^(- 1/2)

y' = ± x / (225 - x²)^(1/2)

y' = ± x / √(225 - x²) ← this is the slope of the tangent line (ℓ₂) to the circle at x

y' = ± x / √(225 - x²) → when: x = 12

y' = ± 12 / √(225 - 144)

y' = ± 12/9

y' = ± 4/3 → you can see on the drawing, that the slope is negative

y' = - 4/3



The slope of the line (ℓ₁) is (3/4).

The slope of the line (ℓ₂) is (- 3/4).

The product of these 2 slopes is (- 1), you can say that the line (ℓ₁) and the line (ℓ₂) are perpendicular.

…and the circles are normal to each other at that point x = 9....Show more

Are you trying to show the slopes are the same?
No, we must prove that the product of the two slopes is -1 
(condition of orthogonality between two lines)

1. 
-(x-6)/(y-17)  @ (12,9) we get m₁ = 3/4
-x/y @  (12,9) we get m₂ = -4/3

2
m₁*m₂ = -1

The two tangent lines are orthogonal this means that the two curves are normal to each other....Show more