(A) Let $(x, y)$ be any point whose distance from $(-3, 0)$ is greater than $2$ units.Using the distance formula,the condition is $\sqrt{(x - (-3))^2

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$\{(x, y) \mid x^2+y^2+6x+5 > 0\}$

(A) Let $(x, y)$ be any point whose distance from $(-3, 0)$ is greater than $2$ units.Using the distance formula,the condition is $\sqrt{(x - (-3))^2 + (y - 0)^2} > 2$.Squaring both sides,we get $(x + 3)^2 + y^2 > 2^2$.Expanding the expression,we have $x^2 + 6x + 9 + y^2 > 4$.Rearranging the terms,we get $x^2 + y^2 + 6x + 9 - 4 > 0$.Thus,the locus is $x^2 + y^2 + 6x + 5 > 0$.

Class 11 Mathematics 10-1.Circle and System of Circles Locus Related Problem

Easy

The locus of all points that are at a distance greater than $2$ units from $(-3, 0)$ is:

TS EAMCET 2019 All TS EAMCET

A
$\{(x, y) \mid x^2+y^2+6x+5 > 0\}$
B
$\{(x, y) \mid x^2+y^2+6x+7 > 0\}$
C
$\{(x, y) \mid x^2+y^2-6x+5 > 0\}$
D
$\{(x, y) \mid x^2+y^2-6x+5 \leq 0\}$

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The locus of all points that are at a distance greater than $2$ units from $(-3, 0)$ is:

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