Question

Two circles of equal radius $a$ cut orthogonally. If their centres are $(2,3)$ and $(5,6)$, then radical axis of these circles passes through the point

• A. $(3a,5a)$
• B. $(2a,a)$
• C. $(a,\frac{5a}{3})$
• D. $(a,a)$

Answer 1

Answer: (C)

$(a,\frac{5a}{3})$

Intersection of circles orthogonally means that the tangents to each of them at their points of intersection are at 90 degrees and passes through the centres of each other.

As radii is same, therefore their radical axis passes through the mid point of both the centres, which is $(\frac{7}{2},\frac{9}{2})$.

Also, by applying pythagoras theorem, $c_1{c}_2=a\sqrt{2}=\sqrt{(5-2{)}^2+(6-3{)}^2}$

$\Rightarrow$ $a=3$

Radical axis is perpendicular to the line joining centres and passes through their mind point, therefore it's equation is given by

$(y-\frac{9}{2})=-1(x-\frac{7}{2})$

=> $x+y=8$

From the above options, option $(a,\frac{5a}{3})$ satisfies the given equation.

Thus the answer is $(a,\frac{5a}{3})$