Question:

Radius $r = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}$ units $\Rightarrow r^2 = (5\sqrt{2})^2 = 50$. [1.0 Mark]
Distance squared $CP^2 = (2a - 11)^2 + (a - 7 - (-9))^2 = 50$. [1.0 Mark]
$(2a - 11)^2 + (a + 2)^2 = 50 \Rightarrow (4a^2 - 44a + 121) + (a^2 + 4a + 4) = 50$. [1.0 Mark]
$5a^2 - 40a + 125 = 50 \Rightarrow 5a^2 - 40a + 75 = 0 \Rightarrow a^2 - 8a + 15 = 0$. [1.0 Mark]
$(a - 3)(a - 5) = 0 \Rightarrow a = 3$ or $a = 5$. Values of $a$ are $3$ and $5$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Calculating radius $r = 5\sqrt{2} \Rightarrow r^2 = 50$: 1.0 Mark
Setting up distance equation $CP^2 = 50$: 1.0 Mark
Expanding to quadratic $5a^2 - 40a + 75 = 0$: 1.0 Mark
Simplifying to $a^2 - 8a + 15 = 0$: 1.0 Mark
Solving for $a = 3, 5$: 1.0 Mark