Question

$A=(x_1,y_1)$ , $B=(x_2,y_2),C=(x_3,y_3)$ then the radical centre of the circles $(x-x_1{)}^2+(y-y_1{)}^2=$ $a^2$
$(x-x_2{)}^2+(y-y_2{)}^2=a^2,(x-x_3{)}^2+(y-y_3{)}^2=a^2$ is the

• A. $C$entroid of $\Delta ABC$
• B. Orthocentre of $\Delta ABC$
• C. Incentre of $\Delta ABC$
• D. $C$ircumcentre of $\Delta ABC$

Answer 1

The correct option is D $C$ircumcentre of $\Delta ABC$

$s_1=(x-x_1{)}^2+(y-y_1{)}^2=a^2$
$s_2=(x-x_2{)}^2+(y-y_2{)}^2=a^2$
$s_3=(x-x_3{)}^2+(y-y_3{)}^2=a^2$
so, radical axis is first two circles is $s_1-s_2=0$
$2(x_2-x_1)x+2(y_2-y_1)y+x_1^2-x_2^2+y_1^2-y_2^2=0$ ----(1)

We observe that $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ satisfy the above equation.

Also, radical axis is perpendicular to the line joining $(x_1,y_1)$ and $(x_2,y_2)$.

So, the above radical axis is a perpendicular bisector of the line joining $(x_1,y_1)$ and $(x_2,y_2)$.

Similarly, the other two radical axes obtained by taking other two circle pairs will also be perpendicular bisector of line joining corresponding centers.

Hence, the radical center is the point of intersection of perpendicular bisectors of the sides of triangle of which $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ are the vertices.

The point satisfies the condition for circumcenter of the triangle.