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10-1.Circle and System of Circles
medium
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
A
$2x - y - 7 = 0,\,2x + y - 9 = 0$
B
$2x + y + 7 = 0,\,2x + y + 9 = 0$
C
$2x + y - 7 = 0,\,\,2x + y + 9 = 0$
D
$2x - y + 7 = 0,\,2x - y + 9 = 0$
Solution
(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 $\frac{{x – 3}}{{3 – 4}} = \frac{{y + 1}}{{ – 1 – 1}}$
$\Rightarrow 2x – y – 7 = 0$.
Standard 11
Mathematics
