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Zielflug [23.3K]
3 years ago
10

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

Mathematics
2 answers:
Vlad [161]3 years ago
7 0

Answer:

Answer is in the picture

ahrayia [7]3 years ago
6 0

Answer:

<h3>#1</h3>

The normal overlaps with the diameter, so it passes through the center.

<u>Let's find the center of the circle:</u>

  • x² + y² + 2gx + 2fy + c = 0
  • (x + g)² + (y + f)² = c + g² + f²

<u>The center is:</u>

  • (-g, -f)

<u>Since the line passes through (-g, -f) the equation of the line becomes:</u>

  • p(-g) + p(-f) + r = 0
  • r = p(g + f)

This is the required condition

<h3>#2</h3>

Rewrite equations and find centers and radius of both circles.

<u>Circle 1</u>

  • x² + y² + 2ax + c² = 0
  • (x + a)² + y² = a² - c²
  • The center is (-a, 0) and radius is √(a² - c²)

<u>Circle 2</u>

  • x² + y² + 2by + c² = 0
  • x² + (y + b)² = b² - c²
  • The center is (0, -b) and radius is √(b² - c²)

<u>The distance between two centers is same as sum of the radius of them:</u>

  • d = √(a² + b²)

<u>Sum of radiuses:</u>

  • √(a² - c²) + √(b² - c²)

<u>Since they are same we have:</u>

  • √(a² + b²) = √(a² - c²) + √(b² - c²)

<u>Square both sides:</u>

  • a² + b² = a² - c² + b² - c² + 2√(a² - c²)(b² - c²)
  • 2c² = 2√(a² - c²)(b² - c²)

<u>Square both sides:</u>

  • c⁴ = (a² - c²)(b² - c²)
  • c⁴ = a²b² - a²c² - b²c² + c⁴
  • a²c² + b²c² = a²b²

<u>Divide both sides by a²b²c²:</u>

  • 1/a² + 1/b² = 1/c²

Proved

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