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

Step-by-step solution

Let S₁ be the circle with centre (2,3) and radius a. Let S₂ be the circle with centre (5,6) and radius a. Then S₁ (x-2)^2+(y-3)^2-a^2=0 and S₂=(x-5)^2+(y-6)^2-a^2=0 Now, radical axis of these circle is given by S₁-S₂=0 aligned & (x-2)^2+(y-3)^2-a^2-(x-5)^2-(y-6)^2+a^2=0 & (4-4 x+9-6 y)-(25-10 x)-(36-12 y)=0 & 13-4 x-6 y-25+10 x-36+12 y=0 & 6 x+6 y-48=0 & x+y= 48 6 =8 & x+y=8 aligned Since, the given circle cut orthogonally, therefore aligned & 2 g₁ g₂+2 f₁ f₂=C₁+C₂ & 2(-2)(-5)+2(-3)(-6)= ((-2)^2+(-3)^2-a^2 ) & + ((