Question

Match the following :
Cirlces Radical centre
$I.x^2+y^2=1$, $a$ $\left)\right(0,0)$
$x^2+y^2-2x=1$,
$x^2+y^2-2y=1$
$II.x^2+y^2-x+3y-3=0,b\left)\right(2,3)$
$x^2+y^2-2x+2y+2=0$
$x^2+y^2+2x+3y-9=0$
$III.x^2+y^2-8x+40=0,c\left)\right(8,-15/2)$
$x^2+y^2-5x+16=0$
$x^2+y^2-8x+16y+160=0$

• A. $a,b,c$
• B. $b,c,a$
• C. $c,a,b$
• D. $a,c,b$

Answer 1

The correct option is A $a,b,c$

Radical centre is a point of intersection of radical axis.
I. $s_1-s_2=0\Rightarrow 2x=0--\left(1\right)$
$s_1-s_3=0\Rightarrow 2y=0--\left(2\right)$
from (1) & (2), radical centre is $(0,0)$
II. $s_1-s_2=0\Rightarrow x+y-5=0--\left(1\right)$
$s_1-s_3=0\Rightarrow -3x+6=0--\left(2\right)$
from (1), & (2), (2,3) is a radical centre.
III. $s_1-s_2=0\Rightarrow -3x+24=0$
$x=8--\left(1\right)$
$s_1-s_3=0\Rightarrow -16y-120=0$
$y=\frac{-30}{4}=-\frac{15}{2}--\left(2\right)$
So, $(8,\frac{-15}{2})$ is a radical centre.