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

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
2:prove that the two circles x²+y²+2ax+c²=0 and x²+y²+2by+c²=0 touch if
\frac{1}{a²}+\frac{1}{b²}=\frac{1}{c²}​

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Ladonna 3 years 2021-08-31T04:21:48+00:00 2 Answers 3 views 0

Answers ( )

  1. diemkieu
    0
    2021-08-31T04:22:56+00:00
    Reply

    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

  2. Euphemia
    0
    2021-08-31T04:23:47+00:00
    Reply

    Answer:

    Answer is in the picture

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )