Question:
The set of all points that are at a distance of at least \(2\) units from \((-3,0)\) is
The set of all points that are at a distance of at least \(2\) units from \((-3,0)\) is
Show Hint
For a point \((x,y)\) to be at least \(r\) units away from \((a,b)\), use
\[
(x-a)^2+(y-b)^2\geq r^2.
\]
Updated On: Jun 24, 2026
- \(\{(x,y)\mid x^2+y^2+6x-7\gt 0\}\)
- \(\{(x,y)\mid x^2+y^2+6x+5\geq 0\}\)
- \(\{(x,y)\mid x^2+y^2+6x+5\lt 0\}\)
- \(\{(x,y)\mid x^2+y^2+6x+7\leq 0\}\)
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The Correct Option is B
Solution and Explanation
Step 1: Use the distance formula.
The distance of a point \((x,y)\) from \((-3,0)\) is \[ \sqrt{(x+3)^2+(y-0)^2} \] Since the distance is at least \(2\), we have \[ \sqrt{(x+3)^2+y^2}\geq 2 \]
Step 2: Square both sides.
\[ (x+3)^2+y^2\geq 4 \] Expanding, \[ x^2+6x+9+y^2\geq 4 \] \[ x^2+y^2+6x+5\geq 0 \]
Step 3: Match with the options.
The required set is \[ \{(x,y)\mid x^2+y^2+6x+5\geq 0\} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\{(x,y)\mid x^2+y^2+6x+5\geq 0\}} \]
The distance of a point \((x,y)\) from \((-3,0)\) is \[ \sqrt{(x+3)^2+(y-0)^2} \] Since the distance is at least \(2\), we have \[ \sqrt{(x+3)^2+y^2}\geq 2 \]
Step 2: Square both sides.
\[ (x+3)^2+y^2\geq 4 \] Expanding, \[ x^2+6x+9+y^2\geq 4 \] \[ x^2+y^2+6x+5\geq 0 \]
Step 3: Match with the options.
The required set is \[ \{(x,y)\mid x^2+y^2+6x+5\geq 0\} \]
Step 4: Final conclusion.
Therefore, \[ \boxed{\{(x,y)\mid x^2+y^2+6x+5\geq 0\}} \]
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