Gradient of normal at point (-2, -3) on circle {x^2} + {y^2} + 2x + 4y + 3 = 0 is

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10-1.Circle and System of Circles

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The gradient of the normal at the point $(-2, -3)$ on the circle ${x^2} + {y^2} + 2x + 4y + 3 = 0$ is

A

$1$

B

$-1$

C

$\frac{3}{2}$

D

$\frac{1}{2}$

Solution

(a) The equation of tangent at point $( – 2,\; – 3)$ to the circle ${x^2} + {y^2} + 2x + 4y + 3 = 0$ is,

$ – 2x – 3y + 1(x – 2) + 2(y – 3) + 3 = 0$

$ \Rightarrow – 2x – 3y + x – 2 + 2y – 6 + 3 = 0$

$ \Rightarrow – x – y – 5 = 0 $

$\Rightarrow x + y + 5 = 0$

or $y = – x – 5$; so, $m = – 1$

Hence, gradient of normal $ = \frac{{ – 1}}{{ – 1}} = 1$.

Standard 11

Mathematics

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