Question

If two circles $2x^2+2y^2-3x+6y+k=0$ and $x^2+y^2-4x+10y+16=0$cut orthogonally, then the value of k is

• A. 41
• B. 14
• C. 4
• D. 1

Answer 1

Answer: (C)

We have given two circles are
$2x^2+2y^2-3x+6y+k=0$
$\Rightarrow x^2+y^2-\frac{3}{2}x+3y+\frac{k}{2}=0$
and $x^2+y^2-4x+10y+16=0$
Since, general equation of circle is
$x^2+y^2+2gx+2fy+c=0$
Therefore, comparing Eqs. (i) and (ii) with Eq. (iii), we get
$g_1=-\frac{3}{4},f_1=\frac{3}{2},c_1=\frac{k}{2}$
and $g_2=-2,f_2=5,c_2=16$
Since, both the circles cut orthogonally.
$\therefore 2\left(g_1{g}_2+f_1{f}_2\right)=c_1+c_2$
$\Rightarrow 2\left(\frac{3}{2}+\frac{15}{2}\right)=\frac{k}{2}+16$
$\Rightarrow 18=\frac{k}{2}+16$
$\Rightarrow \frac{k}{2}=2$
$\Rightarrow k=4$