The equations of the tangents to the circle x | vedclass

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Question

(A) Given the circle equation $x^{2}+y^{2}=13$. Since the abscissa of the points is $x=2$,we substitute this into the circle equation: $2^{2}+y^{2}=13

Vedclass · Accepted Answer

$2x+3y=13, 2x-3y=13$

Vedclass · Answer

(A) Given the circle equation $x^{2}+y^{2}=13$. Since the abscissa of the points is $x=2$,we substitute this into the circle equation: $2^{2}+y^{2}=13$ $4+y^{2}=13$ $y^{2}=9$ $y=\pm 3$. So,the points of contact are $(2, 3)$ and $(2, -3)$. The equation of the tangent to the circle $x^{2}+y^{2}=r^{2}$ at point $(x_{1}, y_{1})$ is given by $xx_{1}+yy_{1}=r^{2}$. For point $(2, 3)$: $2x+3y=13$. For point $(2, -3)$: $2x-3y=13$. Thus,the equations are $2x+3y=13$ and $2x-3y=13$.

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

The equations of the tangents to the circle $x^{2}+y^{2}=13$ at the points whose abscissa is $2$ are:

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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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