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.?

Answer 1

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. 

Answer 2

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