The normal to the circle {x^2} + {y^2} - 3x - | vedclass

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Question

(a) $\frac{{x - {x_1}}}{{{x_1} + g}} = \frac{{y - {y_1}}}{{{y_1} + f}}$

$ \Rightarrow \frac{{x + 3}}{{ - 3 - \frac{3}{2}}} = \frac{{y - 4}}{{4 - 3}

Vedclass · Accepted Answer

$2x + 9y - 30 = 0$

Vedclass · Answer

(a) $\frac{{x - {x_1}}}{{{x_1} + g}} = \frac{{y - {y_1}}}{{{y_1} + f}}$

$ \Rightarrow \frac{{x + 3}}{{ - 3 - \frac{3}{2}}} = \frac{{y - 4}}{{4 - 3}} $

$\Rightarrow 2x + 9y - 30 = 0$.

The normal to the circle ${x^2} + {y^2} - 3x - 6y - 10 = 0$at the point $(-3, 4)$, is

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