Question

The value of k, such that the equation
$2x^2+2y^2-6x+8y+k=0$ represent a point circle , is

• A. 0
• B. 25
• C. $\frac{25}{2}$
• D. $-\frac{25}{2}$

Answer 1

The correct option is C $\frac{25}{2}$

The given equation of circle is

$\Rightarrow$ $2x^2+2y^2-6x+8y+k=0$

$\Rightarrow$ $x^2+y^2-3x+4y+\frac{k}{2}=0$ [ Dividing both sides by $2$ ]

Comparing it with general equation of circle $x^2+y^2+2gx+2fy+c=0,$ we get

$\Rightarrow$ $2g=-3,2f=4$ and $c=\frac{k}{2}$

$\therefore$ $g=\frac{-3}{2}$ and $f=2$

Center $=(-g,-f)=(\frac{3}{2},-2)$

We know,

Radius $\left(r\right)=\sqrt{g^2+f^2-c}$

We know that for point circle radius is $0.$ $r=0$

$\Rightarrow$ $0=\sqrt{{\left(\frac{3}{2}\right)}^2+(-2{)}^2-\frac{k}{2}}$

$\Rightarrow$ $0=\sqrt{\frac{9}{4}+4-\frac{k}{2}}$

$\Rightarrow$ $0=\frac{9}{4}+4-\frac{k}{2}$

$\Rightarrow$ $\frac{k}{2}=\frac{9+16}{4}$

$\Rightarrow$ $\frac{k}{2}=\frac{25}{4}$

$\therefore$ $k=\frac{25}{2}$