The normal to the circle x^2 + y^2 - 3x - 6y | vedclass

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Question

(A) The equation of the circle is $x^2 + y^2 - 3x - 6y - 10 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -\f

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$2x + 9y - 30 = 0$

Vedclass · Answer

(A) The equation of the circle is $x^2 + y^2 - 3x - 6y - 10 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = -\frac{3}{2}$ and $f = -3$.The center of the circle is $(-g, -f) = (\frac{3}{2}, 3)$.The normal to a circle at any point $(x_1, y_1)$ always passes through the center of the circle.The slope of the line passing through the center $(\frac{3}{2}, 3)$ and the point $(-3, 4)$ is $m = \frac{4 - 3}{-3 - \frac{3}{2}} = \frac{1}{-\frac{9}{2}} = -\frac{2}{9}$.The equation of the normal line passing through $(-3, 4)$ with slope $m = -\frac{2}{9}$ is given by $y - y_1 = m(x - x_1)$.$y - 4 = -\frac{2}{9}(x + 3)$$9(y - 4) = -2(x + 3)$$9y - 36 = -2x - 6$$2x + 9y - 30 = 0$.

$A. (7, 2)$	$1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$	$2. -8$
$C. (1, -1)$	$3. 4$
$D. (3, \sqrt{2})$	$4. 0$
	$5. -2\sqrt{2}$

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

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