Question

If the external center of similitude of the circles $x^2+y^2-2x-4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is $Q$, then the value of $Q$ is

• A. $(-4,-3)$
• B. $(4,3)$
• C. $(4,-3)$
• D. $(-4,3)$

Answer 1

Answer: (B)

$(4,3)$

Using section formula, external centre of similitude, is

$Q\equiv (\frac{r_1{x}_2-x_1{r}_2}{r_1{r}_2},\frac{r_1{y}_2-r_2{y}_1}{r_1-r_2})$
From equations,

$r_1=1$, $r_2=2$
$C_1\equiv (1,2)$, $C_2\equiv (-2,1)$
Substituting these values in the above formula, we get
$Q\equiv (\frac{-4}{-1},\frac{-3}{-1})$
$\Rightarrow Q\equiv (4,3)$

Hence, option B.