Find the centre of a circle passing through the points $(6, -6), (3, -7)$ and $(3, 3)$.
Step-by-Step Solution
Key Concept: Let centre $O(x, y)$. $OA^2 = OB^2 = OC^2$. $OB^2 = OC^2 \Rightarrow (x-3)^2 + (y+7)^2 = (x-3)^2 + (y-3)^2 \Rightarrow y^2 + 14y + 49 = y^2 - 6y + 9 \Rightarrow 20y = -40 \Rightarrow y = -2$. Substitute $y = -2$ in $OA^2 = OC^2 \Rightarrow (x-6)^2 + 16 = (x-3)^2 + 25 \Rightarrow x^2 - 12x + 52 = x^2 - 6x + 34 \Rightarrow 6x = 18 \Rightarrow x = 3$. Centre is $(3, -2)$.
Equating $OB^2 = OC^2$: $(y + 7)^2 = (y - 3)^2 \Rightarrow 14y + 49 = -6y + 9 \Rightarrow 20y = -40 \Rightarrow y = -2$. [1.5 Marks]
Equating $OA^2 = OC^2$ with $y = -2$: $(x - 6)^2 + 16 = (x - 3)^2 + 25 \Rightarrow -12x + 52 = -6x + 34 \Rightarrow 6x = 18 \Rightarrow x = 3$. [1.0 Mark]
The centre of the circle is $(3, -2)$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Solving $y = -2$ from $OB^2 = OC^2$: 1.5 Marks
Solving $x = 3$ from $OA^2 = OC^2$: 1.0 Mark
Writing centre $(3, -2)$: 0.5 Mark
Correct Answer: