Question

If $C_{1}$ and $C_{2}$ are the centres of similitude with respect to the circles $x^{2}+y^{2}-14 x+6 y+33=0$ and $x^{2}+y^{2}+30 x-2 y+1=0$, then the equation of the circle with $C_{1} C_{2}$ as diameter is

• A. $2 x^{2}+2 y^{2}+30 x-33 y-17=0$
• B. $2 x^{2}+2 y^{2}-14 x+9 y-13=0$
• C. $2 x^{2}+2 y^{2}-39 x+14 y+74=0$
• D. $2 x^{2}+2 y^{2}-24 x+8 y-5=0$

Answer 1

Answer: (C)

Given equation of circles are
$x^{2}+y^{2}-14 x+6 y+33=0$ ...(i)
and $x^{2}+y^{2}+30 x-2 y+1=0$ ...(ii)
Clearly, circle (i) has centre $O_{1}(7,-3)$ and radius
$r_{1}=\sqrt{49+9-33}=5$
and circle (ii) has centre $O_{2}(-15,1)$
and radius $r_{2}=\sqrt{225+1-1}=15$
We know that the centres of similitude of two circles, whose centres are $A$ and $B$, are the points which divide $A B$ internally and externally in the ratio of radii $r_{a}, r_{b}$.
So, $C_{1}=$ point which divide $O_{1} O_{2}$, internally in the ratio $5: 15$, i.e. $1: 3$
$=\left(\frac{-15+21}{4}, \frac{1-9}{4}\right)=\left(\frac{3}{2},-2\right)$

and $C_{2}=$ Point which divide $O_{1} O_{2}$ externally in the ratio $1: 3$

$=\left(\frac{-15-21}{1-3}, \frac{1+9}{1-3}\right)=\left(\frac{-36}{-2}, \frac{10}{-2}\right)=(18,-5)$
Now, equation of circle with as $C_{1} C_{2}$ diameter is given by
$\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{1}\right)\left(y-y_{2}\right)=0$
$=\left(x-\frac{3}{2}\right)(x-18)+(y+2)(y+5)=0$
$\Rightarrow (2 x-3)(x-18)+2(y+2)(y+5)=0$
$\Rightarrow 2 x^{2}-36 x-3 x+54+2\left(y^{2}+7 y+10\right)=0$
$\Rightarrow 2 x^{2}-39 x+54+ 2 y^{2}+14 y+20=0$
$\Rightarrow 2 x^{2}+2 y^{2}-39 x+14 y+74=0$