Question

The centre of similitude of the two circles $x^2+y^2+4x+2y-4=0$ and $x^2+y^2-4x-2y+4=0$ is

• A. $(4,1)$
• B. $(1,-4)$
• C. $(1,4)$
• D. $(4,2)$

Answer 1

Answer: (D)

$(4,2)$

From the given two equations, we get
$r_1=3$ and $r_2=1$
$C_1=(-2,-1)$ $C_2=(2,1)$
$P(\frac{r_1{x}_2+r_2{x}_1}{r_1+r_2},\frac{r_1{y}_2+r_2{y}_1}{r_1+r_2})$
$P(\frac{4}{4},\frac{2}{4})$
$P(1,\frac{1}{2})$ and $Q(\frac{r_1{x}_2-r_2{x}_1}{r_1-r_2},\frac{r_1{x}_2-r_2{y}_1}{r_1-r_2})$
$Q(\frac{8}{2},\frac{4}{2})$
$Q(4,2)$
So, $Q(4,2)$ is a centre of similitude of two circles.