let C be a circle of radius 5 meters having centre at O.

Given: Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O.

Asked: Then, the length of PQ, in meters, is nearest to



Equation of the circle C:-
\(x^2 + y^2 = 5^2 = 25\)

Equation of line PQ passing through points A(0,4) & B(3,0): -
(y - 4) = (0-4)/(3-0) (x - 0)
y - 4 = -4x/3
4x + 3y = 12

Coordinate P & Q lies on the circle and on the line PQ so they must satisfy equation of the circle C and the line PQ.
\(x^2 + y^2 = 25\)
4x + 3y = 12
y = 4 - 4x/3

\(x^2 + (4 - 4x/3)^2 = 25\)
\(x^2 + 16 + 16x^2/9 - 32x/3 = 25\)
\(9x^2 + 144 + 16x^2 - 96x = 225\)
\(25x^2 - 96x - 81 = 0\)
\(x = \frac{(96 + - \sqrt{96^2 + 4*81*25})}{2*25} = \frac{(96 +- \sqrt{17316})}{50} = \frac{(96 + - 131.59)}{50} = (-.71, 4.55) \)

Point P : -
x = -.71
y = 4- 4x/3 = 4 + 4*.71/3 = 4.95
P = (-.71, 4.95)

Point Q: -
x = 4.55
y = 4 - 4x/3 = 4 - 4*4.55/3 = -2
Q = (4.55, -2)

Length of line segment PQ joining points P(-.71, 4.95) & Q(4.55, -2) : -
\(= \sqrt{(4.55+.71)^2 + (-2-4.95)^2} = \sqrt{75.97} = 8.8\)


IMO C