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# 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.?

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- lenpol7Lv 72 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.

- Steve4PhysicsLv 72 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

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