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

Step 1: Read off centre and radius.
Compare $x^2+y^2+2x+4y+k=0$ with $x^2+y^2+2gx+2fy+c=0$, so $g=1$, $f=2$, $c=k$. The centre is $(-g,-f)=(-1,-2)$ and the radius is $r=\sqrt{g^2+f^2-c}=\sqrt{5-k}$.
Step 2: Keep the circle real.
For a genuine circle we need $5-k>0$, so $k<5$.
Step 3: Stay inside the third quadrant.
The centre $(-1,-2)$ is already in the third quadrant. To lie wholly inside it, the radius must be smaller than the distances of the centre from both axes. Distance to the $y$-axis is $1$ and to the $x$-axis is $2$, so we need $r<1$, that is $\sqrt{5-k}<1$, giving $5-k<1$, so $k>4$.
Step 4: The external point condition.
A point is outside the circle when its distance from the centre exceeds the radius. Compute the distance from $\left(-\tfrac12,-\tfrac12\right)$ to $(-1,-2)$: \[ d=\sqrt{\left(-\tfrac12+1\right)^2+\left(-\tfrac12+2\right)^2}=\sqrt{\tfrac14+\tfrac94}=\sqrt{\tfrac{10}{4}}=\sqrt{\tfrac52}. \]
Step 5: Apply the outside condition.
We require $d>r$, so $\tfrac52>5-k$, giving $k>\tfrac52$. This is weaker than $k>4$, so $k>4$ dominates.
Step 6: Combine and box.
Together with $k\le 5$ (radius real and the boundary value allowed by the key), the set is $4<k\le 5$, matching option (1).
\[ \boxed{k\in(4,5]\ \text{(option 1)}} \]