Question

If $2x^2+2y^2+4x+5y+1=0$ and $3x^2+3y^2+6x-7y+3k=0$ are orthogonal, then value of $k$ is

• A. $\frac{17}{12}$
• B. $\frac{12}{17}$
• C. $\frac{-12}{17}$
• D. $\frac{-17}{12}$

Answer 1

Answer: (D)

Given equation of circle
$2x^2+2y^2+4x+5y+1=0$
$\Rightarrow x^2+y^2+2x+5/2y+1/2=0$...(i)
$3x^2+3y^2+6x-7y+3k=0$
$\Rightarrow x^2+y^2+2x+7/3+k=0$...(ii)
Centre and radius of first circle and second circle
$R_1\to (-1,-5/4),$
$R_1\to \sqrt{1+\frac{25}{16}-1/2}=\frac{\sqrt{16+25-8}}{16}=\sqrt{\frac{33}{16}}$
$C_2\to (-1,7/6),$
$R_2\to \sqrt{1+\frac{49}{36}-k}=\sqrt{\frac{85-36k}{36}}$
Now, condition for orthogonality of two circles
${\left(C_1{C}_2\right)}^2=R_1^2+R_2^2$
$(-1+1{)}^2+(7/6+5/4{)}^2$
$=\frac{33}{16}+\frac{85-36k}{36}$
$0+{\left(\frac{29}{12}\right)}^2=\frac{33}{16}+\frac{85-36k}{36}$
$\frac{841}{144}=\frac{33}{16}+\frac{85-36k}{36}$
$841=297+340-144k$
$144k=637-841=-204$
$k=-\frac{17}{12}$