The equation of the normal to the circle 2x^2 | vedclass

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(C) The given equation of the circle is $2x^2 + 2y^2 - 2x - 5y + 3 = 0$.Dividing by $2$,we get $x^2 + y^2 - x - \frac{5}{2}y + \frac{3}{2} = 0$.The eq

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$x + 2y = 3$

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(C) The given equation of the circle is $2x^2 + 2y^2 - 2x - 5y + 3 = 0$.Dividing by $2$,we get $x^2 + y^2 - x - \frac{5}{2}y + \frac{3}{2} = 0$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 - \frac{1}{2}(x + x_1) - \frac{5}{4}(y + y_1) + \frac{3}{2} = 0$.Substituting $(1, 1)$,we get $x + y - \frac{1}{2}(x + 1) - \frac{5}{4}(y + 1) + \frac{3}{2} = 0$.Multiplying by $4$,we get $4x + 4y - 2x - 2 - 5y - 5 + 6 = 0$,which simplifies to $2x - y - 1 = 0$.The slope of the tangent is $m_t = 2$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.The equation of the normal at $(1, 1)$ is $y - 1 = -\frac{1}{2}(x - 1)$.$2y - 2 = -x + 1$,which gives $x + 2y = 3$.

The equation of the normal to the circle $2x^2 + 2y^2 - 2x - 5y + 3 = 0$ at the point $(1, 1)$ is:

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