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