The line $x - y - 3 = 0$ touches the circle $x^{2} + y^{2} - 4 x + 6 y + 11 = 0$ at a point whose coordinates are

(1, - 2)

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(1, 2)

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(- 1, 2)

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(- 1, - 2)

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Solution

The correct option is A $(1, - 2)$
Equation of the line is $x - y - 3 = 0$ $\Rightarrow x = y + 3$ ...(i)

and equation of the circle is $x^{2} + y^{2} - 4 x + 6 y + 11 = 0$ ...(ii)

Substituting the value of $x$ from Eq,(i) in Eq. (ii), we get

${(y + 3)}^{2} + y^{2} - 4 (y + 3) + 6 y + 11 = 0$

$\Rightarrow y^{2} + 6 y + 9 + y^{2} - 4 y - 12 + 6 y + 11 = 0$

$\Rightarrow 2 y^{2} + 8 y + 8 = 0 \Rightarrow y^{2} + 4 y + 4 = 0$

$\Rightarrow {(y + 2)}^{2} = 0 \Rightarrow y = - 2$

Substituting this value of $y$ in Eq. (i), we get $x = - 2 + 3 = 1.$

Therefore, coordinates of the point $= (1, - 2) .$

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