Question

$x^2+y^2+kx+(1-k)y+5=0$ respresents a circle with radius less than or equal to 5. Then number of integral values of ` $k$' are

• A. $14$
• B. $15$
• C. $16$
• D. $13$

Answer 1

Answer: (C)

$16$

$x^2+y^2+kx+(1-k)y+5=0$
Radius $=\sqrt{{\left(\frac{k}{2}\right)}^2+{\left(\frac{1-k}{2}\right)}^2-5}$
Given, radius $\le$ 5.
$\sqrt{\frac{k^2}{4}+\frac{k^2-2k+1}{4}-5}$ $\le$ 5.
$2K^2-2K-19\le 100$
$2K^2-2K-119\le 0$
$K=(\frac{2-\sqrt{956}}{4},\frac{2+\sqrt{956}}{4})$
$\frac{2+\sqrt{956}}{4}$
$K=-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,8,78$
No. of integer values of $k$ are $16$.