The equations of the normals to the circle {x | vedclass

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Question

(a) The abscissa of point is found by substituting the ordinates and solving for abscissa.

$ \Rightarrow {x^2} - 8x + 15 = 0$

$ \Rightarrow x =

Vedclass · Accepted Answer

$2x - y - 7 = 0,\,2x + y - 9 = 0$

Vedclass · Answer

(a) The abscissa of point is found by substituting the ordinates and solving for abscissa.

$ \Rightarrow {x^2} - 8x + 15 = 0$

$ \Rightarrow x = \frac{{8 \pm \sqrt {64 - 60} }}{2} $

$= \frac{{8 \pm 2}}{2} = 5$ or $3$

i.e., points are $(5,\; - 1)$ and $(3, -1)$.

Normal is given by, $\frac{{x - 5}}{{5 - 4}} = \frac{{y + 1}}{{ - 1 - 1}} $

$\Rightarrow 2x + y - 9 = 0$

and $\frac{{x - 3}}{{3 - 4}} = \frac{{y + 1}}{{ - 1 - 1}}$

$\Rightarrow 2x - y - 7 = 0$.

The equations of the normals to the circle ${x^2} + {y^2} - 8x - 2y + 12 = 0$ at the points whose ordinate is $-1,$ will be

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