In the xy co-ordinate system, a circle C is given by the equation [m]x

Here is the OE

Solution
Step 1: Understand Question Statement
• \(x^2+y^2=36\) is circle C
• Triangle PQR is inscribed in the Circle C.
• Side PQ of the triangle PQR, lies on the x- axis.
• Vertex R of the triangle PQR lies on the line \(y+x-6=0\)
We need to find the area of triangle PQR.

\(Step 2: Define Methodology \)
• Center of C is (0,0) and radius is 6.
• We will find the intersection of the given line and Circle C to get vertex R.
• We will draw the diagram as per the information given in the question.
• Using the properties of triangle and circles, we will find the area PQR.

Step 3: Calculate the final answer
• Let’s find the intersection of C and the given line, substitute the value of y from the equation of the line into the equation of C:
• \(x^2+{(6-x)}^2=36\) ⟹ \(2x(x-6)=0\)
• Substituting the above values of x into equation of line, we get:
• ⟹y-0-6=0 ⟹ y=6 , Hence, the intersection point is (0,6)
• ⟹y-6-6=0 ⟹ y=0 , Hence the intersection point is (6,0)
• Based on above information we get below diagram:


• Since PQR is inscribed in the circle with one side PQ as its diagonal.
• OR, OP and OQ are radii of C.
• Since RO is perpendicular to PQ,
• Area of triangle PQR = \(\frac{1}{2}\ast\ base\ast\ height=\frac{1}{2}\ast\ PQ\ast\ RO=\frac{1}{2}\ast12\ast6=36 \)
Thus, the correct answer is Option C.