Question

If $C_1$ and $C_2$ are the centres of similitude with respect to the circles $x^2+y^2-14x+6y+33=0$ and $x^2+y^2+30x-2y+1=0$, then the equation of the circle with $C_1{C}_2$ as diameter is

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

Answer 1

Answer: (C)

Given equation of circles are
$x^2+y^2-14x+6y+33=0$ ...(i)
and $x^2+y^2+30x-2y+1=0$ ...(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 $AB$ 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
$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$
$=\left(x-\frac{3}{2}\right)(x-18)+(y+2)(y+5)=0$
$\Rightarrow (2x-3)(x-18)+2(y+2)(y+5)=0$
$\Rightarrow 2x^2-36x-3x+54+2\left(y^2+7y+10\right)=0$
$\Rightarrow 2x^2-39x+54+2y^2+14y+20=0$
$\Rightarrow 2x^2+2y^2-39x+14y+74=0$