Let $C$ be a circle of radius $5$ meters having center at $O$. Let $P Q$ 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 $O$. Then, the length of $P Q$, in meters, is nearest to

We have:

Circle C with center O and radius 5 meters

Point A: 4 meters north of O

Point B: 3 meters east of O

Chord PQ passes through both A and B

When a chord passes through two given points, those points lie on the chord. So our chord PQ actually contains the line segment AB.

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Let's place O at the origin of our coordinate system:

O is at (0, 0)

A is at (0, 4) [4 meters north]

B is at (3, 0) [3 meters east]

Using the Pythagorean theorem:

$AB = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ meters

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Since PQ passes through A and B, we need the perpendicular distance from O to line AB.

We'll use the area formula for triangle OAB in two ways:

Using base and height:

Base AB = 5 meters

Height = perpendicular distance from O to AB = d

Area = $\tfrac{1}{2} \times 5 \times d$

Using two sides and included angle:

We can use OA = 4 and OB = 3 as two sides

Since ∠AOB = 90° (A is north, B is east), sin(90°) = 1

Area = $\tfrac{1}{2} \times 4 \times 3 \times 1 = 6$

Setting both equal:

$\tfrac{1}{2} \times 5 \times d = 6$

$d = \tfrac{12}{5} = 2.4$ meters

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In any circle, if we draw a perpendicular from the center to a chord, it bisects the chord. So if D is the foot of perpendicular from O to chord PQ, then PD = DQ.

In right triangle ODP:

OP = 5 (radius of circle)

OD = 2.4 (perpendicular distance we just found)

PD = ? (half the chord length)

Using Pythagorean theorem:

$PD^2 + OD^2 = OP^2$

$PD^2 + (2.4)^2 = 5^2$

$PD^2 + 5.76 = 25$

$PD^2 = 19.24$

$PD = \sqrt{19.24} = 4.4$ meters

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Since the perpendicular from center bisects the chord:

$PQ = 2 \times PD = 2 \times 4.4 = 8.8$ meters

The length of chord PQ is 8.8 meters.

This method works for any chord passing through given points inside a circle. The key insights are using the area method to find the perpendicular distance and applying the chord property that the perpendicular from center always bisects the chord.