Question

If circle \[ x^2+y^2+2x+4y+k=0 \] lies totally inside third quadrant and point \[ \left(-\frac12,-\frac12\right) \] lies outside circle, then set of all real values of \(k\) is

• A. \((4,5]\)
• B. \((\frac12,4]\)
• C. \((\frac52,5)\)
• D. \((5,6]\)

Hint:

For circle lying completely inside a quadrant, radius must be smaller than distance of center from both coordinate axes.

Answer 1

The Correct Option is A

Concept: Circle equation: \[ x^2+y^2+2gx+2fy+c=0 \] Center \[ (-g,-f) \] Radius \[ r=\sqrt{g^2+f^2-c} \]

Step 1: Find center and radius.
Comparing: \[ g=1,\qquad f=2 \] Center \[ (-1,-2) \] Radius \[ r=\sqrt{5-k} \]

Step 2: Condition circle inside third quadrant.
Need radius smaller than distances from axes. Nearest distance to x-axis: \[ 2 \] Nearest distance to y-axis: \[ 1 \] Thus \[ r& lt;1 \] \[ \sqrt{5-k}& lt;1 \] \[ k& gt;4 \]

Step 3: External point condition.
Distance from point to center: \[ d= \sqrt{ \left(-\frac12+1\right)^2+ \left(-\frac12+2\right)^2 } \] \[ = \sqrt{\frac14+\frac94} = \sqrt{\frac52} \] Point outside means \[ d& gt;r \] \[ \frac52& gt;5-k \] \[ k& gt;\frac52 \] Combine both conditions. Also radius real: \[ k\le5 \] Hence \[ 4& lt;k\le5 \] Thus \[ \boxed{(4,5]} \]