Question

If $y=2x+K$ is a diameter to the circle $2(x^2+y^2)+3x+4y-1=0$, then $K$ equals

• A. $0$
• B. $1$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Consider the given equation of the circle.

$2(x^2+y^2)+3x+4y-1=0$

$x^2+y^2+\frac{3}{2}x+2y-\frac{1}{2}=0$

General equation of circle is,

$x^2+y^2+2gx+2fy+c=0$

Therefore,

$2g=\frac{3}{2}\Rightarrow g=\frac{3}{4}$

$2f=2\Rightarrow f=1$

$c=-\frac{1}{2}$

We know that centre of the circle,

$C(-g,-f)$

$C(-\frac{3}{4},-1)$

Since, the diameter of this circle is,

$y=2x+k$

Therefore, the centre of this circle lies on the diameter,

$-1=2\times -\frac{3}{4}+k$

$-1=-\frac{3}{2}+k$

$k=\frac{1}{2}$

Hence, the value of $k$ is $\frac{1}{2}$.