From a point P(3, 3) on the circle `x^(2)+y^(2)=18` two chords PQ and PR each of he length 2 units are drawn on this circle. Then, the value of the length PM is equal to (where, M is the midpoint of the line segment joining Q and R)

Options

A. `(1)/(3sqrt2)" units"`
B. `(1)/(2)" units"`
C. `(sqrt2)/(3)" units"`
D. `(4)/(9)" units"`

Answer

To solve the problem step by step, we will follow the geometric and algebraic approach to find the length of PM, where M is the midpoint of the line segment joining points Q and R on the circle.

### Step 1: Verify Point P on the Circle
First, we need to confirm that the point P(3, 3) lies on the circle defined by the equation \(x^2 + y^2 = 18\).

\[ 3^2 + 3^2 = 9 + 9 = 18 \]

Since this holds true, point P is indeed on the circle.

**Hint:** Always verify if the given point lies on the circle by substituting its coordinates into the circle's equation.

### Step 2: Determine the Circle's Radius
The equation of the circle is \(x^2 + y^2 = 18\). The radius \(r\) can be calculated as:

\[ r = \sqrt{18} = 3\sqrt{2} \]

**Hint:** The radius of a circle can be found by taking the square root of the constant term in the circle's equation.

### Step 3: Understand the Geometry
From point P, two chords PQ and PR are drawn, each of length 2 units. Let O be the center of the circle (0, 0). The distances from O to P, Q, and R are equal to the radius \(3\sqrt{2}\).

**Hint:** Visualize the problem by sketching the circle and marking the points P, Q, and R.

### Step 4: Apply the Cosine Rule in Triangle OQP
In triangle OQP, we can apply the cosine rule to find the angle \( \theta \) at point O:

\[ OQ^2 = OP^2 + PQ^2 - 2 \cdot OP \cdot PQ \cdot \cos(\theta) \]

Substituting the known values:

\[ (3\sqrt{2})^2 = (3\sqrt{2})^2 + 2^2 - 2 \cdot (3\sqrt{2}) \cdot 2 \cdot \cos(\theta) \]

This simplifies to:

\[ 18 = 18 + 4 - 12\sqrt{2} \cos(\theta) \]

Rearranging gives:

\[ 12\sqrt{2} \cos(\theta) = 4 \implies \cos(\theta) = \frac{1}{3\sqrt{2}} \]

**Hint:** The cosine rule relates the lengths of the sides of a triangle to the cosine of one of its angles.

### Step 5: Find Length PM
In triangle QMP, we can use the cosine of angle \( \theta \) to find PM:

\[ \cos(\theta) = \frac{PM}{QP} \]

Substituting the known values:

\[ \frac{1}{3\sqrt{2}} = \frac{PM}{2} \]

Solving for PM gives:

\[ PM = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} \]

**Hint:** Use the relationship between the sides of the triangle and the cosine of the angle to find the desired length.

### Final Answer
The length of PM is:

\[ \boxed{\frac{\sqrt{2}}{3}} \]