Discussion Forum

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 

Answer:

Explanation:

Let  $S_{1}$ be the circle with centre (2,3) and rradius a.

Let $S_{2}$ be the circle  with centre (5,6) and radius a

Then $S_{1}=(x-2)^{2}+(y-3)^{2}-a^{2}=0$

and $S_{2}=(x-5)^{2}+(y-6)^{2}-a^{2}=0$

 Now, radical axis of these circle is given by

$S_{1}-S_{2}$=0

  $(x-2)^{2}+(y-3)^{2}-a^{2}-(x-5)^{2}-(y-6)^{2}+a^{2}=0$

 $\Rightarrow$     $(4-4x+9-6y)-(25-10x)-(36-12y)=0$

 $\Rightarrow$  $13-4x-6y-25+10x-36+12y=0$

 $\Rightarrow$  $6x+6y-48=0$

 $\Rightarrow$    $x+y= \frac{48}{6}=8$

 $\Rightarrow$       $x+y=8$  ........(i)

 Since, the given  circle cut  othogonally, therefore

 $2g_{1}g_{2}+2f_{1}f_{2}=c_{1}+c_{2}$

$\Rightarrow$    $2(-2)(-5)+2(-3)(-6)=((-2)^{2}+(-3)^{2}-a^{2})+((-5)^{2}+(-6)^{2}-a^{2})$

  $[\because radius=\sqrt{g^{2}+f^{2}+c}\therefore a^{2}=g^{2}+f^{2}-c \Rightarrow c=g^{2}+f^{2}-a^{2}]$

$\Rightarrow$   $20+36=(13-a^{2})+(61-a^{2})$

$\Rightarrow$    $56=74-2a^{2}$

$\Rightarrow$    $2a^{2}=74-56=18$

$\Rightarrow$   $a^{2}=9$

$\Rightarrow$   a=3     $[ \because$ radius can't be negative ]

Clearly, option (c)  satisfy the Eq.(i).