If $A$ and $B$ are the centres of similitude with respect to the circles $x^2+y^2-14 x+6 y+33=0$ and…

If $A$ and $B$ 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 midpoint of AB is
  1. $\left(\frac{7}{3}, \frac{4}{5}\right)$
  2. $\left(\frac{3}{2}, \frac{1}{5}\right)$
  3. $\left(\frac{39}{2}, \frac{-7}{4}\right)$
  4. $\left(\frac{39}{4}, \frac{-7}{2}\right)$

Solution

$\begin{aligned} & S_1 \equiv x^2+y^2-14 x+6 y+33=0 \\ & C_1=(7,-3), r_1=\sqrt{49+3-33}=5 \\ & S_2 \equiv x^2+y^2+30 x-2 y+1=0 \\ & C_2=(-15,1), r_2=\sqrt{225+1-1}=15 \end{aligned}$
Let $A$ divides $C_1 C_2$ in $r_1: r_2$ internally $\Rightarrow A=\left(\frac{21-15}{4}, \frac{1-9}{4}\right)=\left(\frac{3}{2},-2\right)$
Let $B$ divides $C_1 C_2$ in $r_1: r_2$ externally $B=\left(\frac{-21-15}{1-3}, \frac{1+9}{1-3}\right)=(18,-5)$
Midpoint of $A B=\left(\frac{18+\frac{3}{2}}{2}, \frac{-5-2}{2}\right)=\left(\frac{39}{4}, \frac{-7}{2}\right)$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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