if 0 is the center of a circle of radius r,Ac and BC are chords of lengths(r-2) and (r+16) respectively then what oculd be the radius of the circle

• In
  Δ
  A
  O
  P
  , by Pythagoras theorem
  
  r
  2
  =
  3
  2
  +
  (
  1
  +
  x
  )
  2
  ....(1)
  
  In
  Δ
  C
  O
  Q
  ,
  r
  2
  =
  4
  2
  +
  x
  2
  ...(2)
  
  e
  q
  n
  (
  1
  )
  −
  e
  q
  n
  (
  2
  )
  O
  =
  9
  −
  16
  +
  (
  1
  +
  x
  )
  2
  −
  x
  2
  
  ⇒
  2
  x
  =
  6
  →
  x
  =
  3
  
  By, eqn (1)
  r
  2
  =
  9
  +
  16
  ⇒
  r
  2
  =
  25
  ⇒
  r
  =
  5
  c
  m
• 8 Months agoHelpfull: Yes(0) No(1)
• Step 1: Set up the equation with equal perpendicular distances
  From the previous setup, we have the following two equations:
  
  𝑑
  1
  2
  +
  (
  𝑟
  −
  2
  )
  2
  4
  =
  𝑟
  2
  d
  1
  2
  ​
  +
  4
  (r−2)
  2
  
  ​
  =r
  2
  
  𝑑
  2
  2
  +
  (
  𝑟
  +
  16
  )
  2
  4
  =
  𝑟
  2
  d
  2
  2
  ​
  +
  4
  (r+16)
  2
  
  ​
  =r
  2
  
  Since
  𝑑
  1
  =
  𝑑
  2
  d
  1
  ​
  =d
  2
  ​
  , we can equate the expressions for the perpendicular distances:
  
  (
  𝑟
  −
  2
  )
  2
  4
  =
  (
  𝑟
  +
  16
  )
  2
  4
  4
  (r−2)
  2
  
  ​
  =
  4
  (r+16)
  2
  
  ​
  
  Step 2: Solve the equation
  To simplify, multiply both sides by 4 to eliminate the denominators:
  
  (
  𝑟
  −
  2
  )
  2
  =
  (
  𝑟
  +
  16
  )
  2
  (r−2)
  2
  =(r+16)
  2
  
  Now expand both sides:
  
  𝑟
  2
  −
  4
  𝑟
  +
  4
  =
  𝑟
  2
  +
  32
  𝑟
  +
  256
  r
  2
  −4r+4=r
  2
  +32r+256
  Cancel
  𝑟
  2
  r
  2
  from both sides:
  
  −
  4
  𝑟
  +
  4
  =
  32
  𝑟
  +
  256
  −4r+4=32r+256
  Step 3: Solve for
  𝑟
  r
  Now, solve for
  𝑟
  r:
  
  −
  4
  𝑟
  −
  32
  𝑟
  =
  256
  −
  4
  −4r−32r=256−4
  −
  36
  𝑟
  =
  252
  −36r=252
  𝑟
  =
  252
  −
  36
  =
  −
  7
  r=
  −36
  252
  ​
  =−7
• 11 Days agoHelpfull: Yes(0) No(0)