Question

If straight line $y=2x+k$ cuts the circle $4x^2+4y^2-4x-8y-15=0$ exactly two real distinct then number integral values of $k$ are :

• A. $11$
• B. $10$
• C. $9$
• D. $8$

Answer 1

The correct option is C $9$

we can say that the line

$y=2x+k$ is secant for circle.

$\begin{array}{cc}\therefore \text{solving line and circle together.}& \\ & \Rightarrow 4x^2+4(2x+k{)}^2-4x-8(2x+k)-15=0\\ \Rightarrow & 4x^2+4(4x^2+4xk+k^2)-4x-16x-8k-15=0\\ \Rightarrow & 20x^2+16kx+4k^2-20x-8k-15=0\\ & \Rightarrow 20x^2+16kx-20x+4k^2-8k-15=0\\ & \Rightarrow 20x^2+x(16k-20)+4k^2-8k-15=0\\ & \therefore \text{line cuts circle at two distinct points}\\ & \text{so we nued}D>0\text{for this quadratic in}x\end{array}$

$(16k-20{)}^2-4(20)(4k^2-8k-15)>0$

$\Rightarrow 16(4k-5{)}^2-80(4k^2-8k-15)>0$

$\Rightarrow 16(16k^2+25-40k)-80(4k^2-8k-15)>0$

$\Rightarrow 16k^2+25-40k-5(4k^2-8k-15)>0$

$\Rightarrow 16k^2+25-4\varphi k-20k^2+40k+75>0$

$\Rightarrow -4k^2+100>0$

$\Rightarrow -4k^2>-100$

$\begin{array}{cc}\Rightarrow & 4k^2<100\\ \Rightarrow & k^2<25\\ \therefore & -5<K<5\end{array}$

so passible integer values of $k$ are $-4,-3,-2,-1,0,1,2,3,4$

i.e there are nine values

$\therefore$ Answer option (c)