The equation of the circle having normal at $(3, 3)$ as $y = x$ and passing through $(2, 2)$ is:

x^{2} + y^{2} - 3 x - 7 y + 12 = 0

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x^{2} + y^{2} - 4 x - 6 y + 12 = 0

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x^{2} + y^{2} - 6 x - 4 y + 12 = 0

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x^{2} + y^{2} - 5 x - 5 y + 12 = 0

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Solution

The correct option is D $x^{2} + y^{2} - 5 x - 5 y + 12 = 0$
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 2 x + 2 y \frac{d y}{d x} + 2 g + 2 f \frac{d y}{d x} = 0. \frac{d y}{d x} = \frac{- y - x}{y + f .} \frac{d y}{d x} a t (3, 3) . \Rightarrow \frac{d y}{d x} = \frac{- y - 3}{3 + f} S l o p e o f n o r m a l a t t h i s p o i n t = 1 \Rightarrow - \frac{(g + 3)}{f + 3} (1) = - 1 \Rightarrow g + 3 = f + 3. \Rightarrow g = f . \Rightarrow x^{2} + y^{2} + 2 g x + 2 f y + c = 0 C i r c l e p a s s e s t h r o u g h (3, 3) & (2, 2) . \Rightarrow g + g + b g + b g + c = 0 \Rightarrow 12 g = - 18 - c . \Rightarrow 4 + 4 + 4 g + 4 g + c = 0. 8 g = - 8 - c . - 12 g = - 18 - c - 4 g = 10 g = \frac{- 5}{2} = f . C = - 8 (g + 1) = - 8 (\frac{- 5}{2} + 1) = - 8 \times \frac{- 3}{2} = + 12 ∴ E q u a t i o n : x^{2} + y^{2} - 5 x - 5 y + 12 = 0. A n s .$

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