Question

The set of values of \(k\), for which the circle \(C : 4x^2 + 4y^2 – 12x + 8y + k = 0\) lies inside the fourth quadrant and the point \(\bigg(1, -\frac{1}{3}\bigg)\) lies on or inside the circle \(C\), is

• A. An empty set
• B. \(\bigg(6/\frac{65}{9}\bigg)\)
• C. \(\bigg(\frac{80}{9},10\bigg)\)
• D. \(\bigg(9,\frac{92}{9}\bigg)\)

Answer 1

The Correct Option is D

 To find the set of values of \( k \) for which the circle \( C: 4x^2 + 4y^2 - 12x + 8y + k = 0 \) lies inside the fourth quadrant, and the point \(\bigg(1, -\frac{1}{3}\bigg)\) lies on or inside the circle, follow these steps:

1. Rewrite the equation of the circle in standard form. The given circle equation is: \(4x^2 + 4y^2 - 12x + 8y + k = 0\).
2. Divide the entire equation by 4 to simplify: \(x^2 + y^2 - 3x + 2y + \frac{k}{4} = 0\).
3. Complete the square to express the equation in standard circle form:
   • For \( x \): \(x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4}\)
   • For \( y \): \(y^2 + 2y = (y + 1)^2 - 1\)
4. Combine and simplify: 

\[(x - \frac{3}{2})^2 + (y + 1)^2 = \frac{9}{4} + 1 - \frac{k}{4} \]\]

1. Further simplify to find the radius: 

\[(x - \frac{3}{2})^2 + (y + 1)^2 = \frac{13}{4} - \frac{k}{4} \]\]

1. Identify the center and radius: \(Center = \left(\frac{3}{2}, -1\right)\), Radius = \(\sqrt{\frac{13}{4} - \frac{k}{4}}\).
2. For the circle to lie entirely in the fourth quadrant, the center must satisfy:
   • \(\frac{3}{2} > 0\), which always holds true.
   • \(-1 < 0\), which also always holds true.
   • The radius must be less than or equal to 1 (since the center is \(\frac{3}{2}\) and this plus 1 lies within the bounds): \[ \sqrt{\frac{13}{4} - \frac{k}{4}} \leq 1 \] Squaring both sides: \[ \frac{13}{4} - \frac{k}{4} \leq 1 \] Solving gives: \[ \frac{13}{4} - 1 \leq \frac{k}{4} \] \[ \frac{9}{4} \leq \frac{k}{4} \] Simplifying: \[ k \geq 9 \]
3. For the point \(\bigg(1, -\frac{1}{3}\bigg)\) to lie on or inside the circle: \[ \left(1 - \frac{3}{2}\right)^2 + \left(-\frac{1}{3} + 1\right)^2 \leq \frac{13}{4} - \frac{k}{4} \] \] Simplify: \[ \left(-\frac{1}{2}\right)^2 + \left(\frac{2}{3}\right)^2 \leq \frac{13}{4} - \frac{k}{4} \] \] \[ \frac{1}{4} + \frac{4}{9} \leq \frac{13}{4} - \frac{k}{4} \] Convert to a common denominator: \[ \frac{9}{36} + \frac{16}{36} \leq \frac{13}{4} - \frac{k}{4} \] \[ \frac{25}{36} \leq \frac{13}{4} - \frac{k}{4} \] Solving gives: \[ 1.083333...\leq 3.25 - \frac{k}{4} \] \[ \frac{k}{4} \leq 2.1666666... \] \[ k \leq \frac{92}{9} \]
4. Therefore, combining both conditions, the set of values for \( k \) is: \[ 9 \leq k \leq \frac{92}{9} \] Which corresponds to the option \(\bigg(9,\frac{92}{9}\bigg)\).