Answer:
#1
The normal overlaps with the diameter, so it passes through the center.
Let's find the center of the circle:
- x² + y² + 2gx + 2fy + c = 0
- (x + g)² + (y + f)² = c + g² + f²
The center is:
Since the line passes through (-g, -f) the equation of the line becomes:
- p(-g) + p(-f) + r = 0
- r = p(g + f)
This is the required condition
#2
Rewrite equations and find centers and radius of both circles.
Circle 1
- x² + y² + 2ax + c² = 0
- (x + a)² + y² = a² - c²
- The center is (-a, 0) and radius is √(a² - c²)
Circle 2
- x² + y² + 2by + c² = 0
- x² + (y + b)² = b² - c²
- The center is (0, -b) and radius is √(b² - c²)
The distance between two centers is same as sum of the radius of them:
Sum of radiuses:
Since they are same we have:
- √(a² + b²) = √(a² - c²) + √(b² - c²)
Square both sides:
- a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
- 2c² = 2√(a² - c²)(b² - c²)
Square both sides:
- c⁴ = (a² - c²)(b² - c²)
- c⁴ = a²b² - a²c² - b²c² + c⁴
- a²c² + b²c² = a²b²
Divide both sides by a²b²c²:
Proved
