Question

lf $(0,\frac{5}{2})$ is a centre of similitude for the circles $x^2+y^2+6x-2y+1=0$ and $x^2+y^2-2x-6y+9=0$ then the length of the common tangent of the circles through it is

• A. $6$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

From Point (1),
$n^2+y^2+6x-2y+1=0$
$r_1=3$
$C_1\equiv (-3,1)$
$x^2+y^2-2x-6y+9=0$
$r_2=1$
$C_2\equiv (1,3)$

From Point (11),
$P\equiv (0,\frac{5}{2})$
& $Q\equiv (\frac{6}{2},\frac{8}{2})$
$Q\equiv \left(3.4\right)$
P is intersection of internal common tangent
So,
length of tangent $=\sqrt{S_1}+\sqrt{S_2}$
$=\sqrt{\frac{9}{4}}+\sqrt{\frac{1}{4}}$
$=\frac{3}{2}+\frac{1}{2}$
$=2$