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Step-by-Step Solution
Step 1: Rewrite the Circle Equation in Standard Form
The given circle is:
36x² + 36y² − 108x + 120y + C = 0
Divide throughout by 36 to simplify:
$x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$
This is of the form:
$x^2 + y^2 + 2gx + 2fy + c = 0$
Comparing terms, we get:
$-3 = 2g \implies g = -\frac{3}{2}$
$\frac{10}{3} = 2f \implies f = \frac{10}{6} = \frac{5}{3}$
$c = \frac{C}{36}$
Step 2: Determine the Centre and Radius
The centre of the circle is $(-g, -f) = \left(\frac{3}{2}, -\frac{5}{3}\right)$. The radius $r$ is given by:
$r = \sqrt{g^2 + f^2 - c}
= \sqrt{\left(-\frac{3}{2}\right)^2 + \left(\frac{5}{3}\right)^2 - \frac{C}{36}}
= \sqrt{\frac{9}{4} + \frac{25}{9} - \frac{C}{36}}.$
Simplifying inside the square root:
$\frac{9}{4} = \frac{81}{36}, \quad \frac{25}{9} = \frac{100}{36}.$
Thus $r = \sqrt{\frac{81}{36} + \frac{100}{36} - \frac{C}{36}}
= \sqrt{\frac{181 - C}{36}}
= \frac{\sqrt{181 - C}}{6}.$
Step 3: Ensure the Circle Does Not Touch or Intersect the Coordinate Axes
The centre has x-coordinate $\frac{3}{2}$. For the circle not to intersect or touch the y-axis, its radius must be strictly less than the distance from the centre to the y-axis (which is $\frac{3}{2}$). Hence:
$r < \frac{3}{2}
\implies \frac{\sqrt{181 - C}}{6} < \frac{3}{2}
\implies \sqrt{181 - C} < 9
\implies 181 - C < 81
\implies C > 100.
(This arises from the requirement that the entire circle lie to the right of the y-axis. Similarly, conditions about the x-axis are also satisfied with the same inequality here because of how the centre’s coordinates and radius work out.)
Step 4: Check the Condition That (2, −1) Lies Inside the Circle
We need the point (2, −1) to satisfy the inequality for the interior of the circle, i.e., plugging into the circle’s equation must yield a negative value.
Substitute (x, y) = (2, −1) into
$x^2 + y^2 - 3x + \frac{10}{3}y + \frac{C}{36} = 0$:
$2^2 + (-1)^2 - 3(2) + \frac{10}{3}(-1) + \frac{C}{36} < 0 \\
4 + 1 - 6 - \frac{10}{3} + \frac{C}{36} < 0.
Combine the constants:
$5 - 6 - \frac{10}{3} = -1 - \frac{10}{3} = -\frac{3}{3} - \frac{10}{3} = -\frac{13}{3}.$
So the inequality becomes:
$-\frac{13}{3} + \frac{C}{36} < 0
\implies \frac{C}{36} < \frac{13}{3}
\implies C < 156.
Step 5: Combine Both Conditions
From the two inequalities, we have:
$C > 100$
$C < 156$
Hence, the required range for $C$ is:
$\boxed{100 < C < 156}$
Final Answer
The correct answer is:
100 < C < 156
