The points B(6, p)and C(6, q) lie on the circle x ^2 + y ^2 − 10x − 6y + 30 = 0 where the value of p < q. Find p and q.?

2 Answers

Relevance
  • 2 months ago

    Since they lie on the circle , they must satisfy the equation. 

    Hence 

    6^2 + p^2 - 10(6) - 6p + 30 = 0 

    p^2 - 6p + 6 = 0 

    Complete the Square 

    p^2 -6p = - 6 

    ( p - 3)^2 - (3)^2 = -6 

    ( p - 3)^2 - 9 = -6 

    ( p - 3)^2 = 3 

    Square root 

    p - 3 = +/-sqrt(3) 

    p = 3 - sqrt(3) = 1.2679...

    Hence q = 3 + sqrt(3) = 4.7320...

    Hence p < q  As required. 

  • 2 months ago

    x² + y² − 10x − 6y + 30 = 0

    Using x=6 gives:

    6² + y² − 10*6 − 6y + 30 = 0

    y² – 6y + 6 = 0

    Solving with the quadratic formula gives

    y = 3±√3

    p and q are the y-values when x=6:

    p = 3-√3

    q = 3+√3

Still have questions? Get answers by asking now.