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

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(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightar

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$(-2,-16)$

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(D) The given equation of the circle is $x^2+y^2-6x+8y-144=0$. Comparing this with the general equation $x^2+y^2+2gx+2fy+c=0$,we get $2g = -6 \Rightarrow g = -3$ and $2f = 8 \Rightarrow f = 4$. The centre of the circle is $(-g, -f) = (3, -4)$. We know that the normal to a circle at any point always passes through its centre. Let the point where the normal meets the circle again be $(x_1, y_1)$. Since the centre $(3, -4)$ is the midpoint of the chord connecting $(8, 8)$ and $(x_1, y_1)$,we have: $\frac{x_1+8}{2} = 3$ $\Rightarrow x_1+8 = 6$ $\Rightarrow x_1 = -2$ $\frac{y_1+8}{2} = -4$ $\Rightarrow y_1+8 = -8$ $\Rightarrow y_1 = -16$ Thus,the required point is $(-2, -16)$.

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

The normal to the circle given by $x^2+y^2-6x+8y-144=0$ at $(8,8)$ meets the circle again at the point

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Match the points on the curve $2y^2 = x + 1$ with the slopes of the normals at those points and choose the correct answer.
$A. (7, 2)$ $1. -4\sqrt{2}$
$B. (0, 1/\sqrt{2})$ $2. -8$
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The line $x + y = 2$ is tangent to the curve $x^2 = 3 - 2y$ at its point

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