Question

The number of integer values of k for which the equation $x^2+y^2+(k-1)x-ky+5=0$ represents a circle whose radius cannot exceed 3, is:

• A. 10
• B. 11
• C. 4
• D. 5

Answer 1

The correct option is C 4

$x^2+y^2+(k-1)x-ky+5=0\to \left(1\right)$

$\therefore$general equation of circle is,

$x62+y^2+gx+2fy+c=0centre=(-g,-f),radius=\sqrt{g^2+f^2-c}$

In equation $\left(1\right)$

$\therefore (-g,-f)=[\frac{-(k-1)}{2},\frac{k}{2}],c=5radius=\sqrt{g^2+f^2-c}radius=\sqrt{\frac{(k-1{)}^2}{4}+\frac{k^2}{4}-5}\therefore \frac{(k-1{)}^2}{4}+\frac{k^2}{4}-5>0\Rightarrow 2k^2-2k-19>0\to \left(1\right)$

Given radius$\le 3\therefore \sqrt{\frac{(k-1{)}^2}{4}+\frac{k^2}{4}-5}\le 3\frac{(k-1{)}^2}{4}+\frac{k^2}{4}-5\le 9\Rightarrow 2k^2-2k-55\le 0\to \left(2\right)$

Solving $\left(1\right)\&\left(2\right)2k^2-2k-19>0\Rightarrow [k-\frac{(1+\sqrt{39})}{2}(k-\left(\frac{1-\sqrt{39}}{2}\right)\left)\right]>0\Rightarrow K\in (-\infty ,\frac{1-\sqrt{39}}{2})\cup (\frac{1+\sqrt{39}}{2},\infty )\to \left(fig1\right)$

$2k^2-2k-55\le 0$

$\Rightarrow [k-\frac{(1+\sqrt{III})}{2}(k-\frac{1-\sqrt{III}}{2}\left)\right)]\le 0\Rightarrow K\in [\frac{1-\sqrt{III}}{2},\frac{1+\sqrt{III}}{2}]\to (fig2)$

By wavy curve method

$\therefore K\in [\frac{1-\sqrt{III}}{2},\frac{1-\sqrt{III}}{2}]\cup [\frac{1+\sqrt{39}}{2},\frac{1+\sqrt{III}}{2}]\therefore K\in [-4.76,-2.622]\cup [3.62,5.76]$

$\therefore$ Integral values of $k$ possible are $-4,-3,-2,4$

$\therefore$ Only $4$ integral values of $k$ are possible.