Question

1:Under what condition will the line px+py+r=0 mat be a normal to the circke x²+y²+2gx+2fy+c=0

Answer 1

Answer:

Answer is in the picture

Answer 2

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:

• (-g, -f)

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:

• d = √(a² + b²)

Sum of radiuses:

• √(a² - c²) + √(b² - c²)

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:

• 1/a² + 1/b² = 1/c²

Proved