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Coordinate Geometry
NCERT Exemplar Ch 07
CBSE_NCERT_EXEMPLAR_CH07
Grade 10

Question:

The center of a circle is $C(2a, a - 7)$. Find the value of $a$ if the circle passes through the point $P(11, -9)$ and has a diameter of length $10\sqrt{2}$ units.

Step-by-Step Solution

Key Concept: Radius $r = \dfrac{\text{Diameter}}{2} = 5\sqrt{2}$. Set distance $CP^2 = r^2 = 50$.
Stepwise Solution:

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]

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

Correct Answer:
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