Question

Two circles of equal radius ${}^{\prime }{a}^{\prime }$ 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. $\left(a,\frac{5a}{3}\right)$
• D. (a, a)

Answer 1

Answer: (C)

Let $S_1$ be the circle with centre $(2,3)$ and radius a.
Let $S_2$ be the circle with centre $(5,6)$ and radius $a$.
Then $S_1\equiv (x-2{)}^2+(y-3{)}^2-a^2=0$
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 orthogonally, therefore
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
$\Rightarrow 2(-2)(-5)+2(-3)(-6)=\left((-2{)}^2+(-3{)}^2-a^2\right)$
$+\left((-5{)}^2+(-6{)}^2-a^2\right)$
$[\because \text{ radius }=\sqrt{g^2+f^2-c}\therefore a^2=g^2+f^2-c\Rightarrow .$
$C=g^2+f^2-a^2]$
$\Rightarrow 20+36=(13-a^2)+(61-a^2)$
$\Rightarrow 56=74-2a^2.$
$\Rightarrow 2a^2=74-56=18$
$\Rightarrow a^2=9$
$\Rightarrow a=3[\because$ radius can't be negative $]$
Clearly, option (c) satisfy the Eq. (i)