Q99 / 160 · TS EAMCET 2017 · Mathematics
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$(3 a, 5 a)$
- B$(2 a, a)$
- Correct answerC$\left(a, \frac{5 a}{3}\right)$
- D$(a, a)$
Solution
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 $S_1-S_2=0$
$$
\begin{aligned}
& (x-2)^2+(y-3)^2-a^2-(x-5)^2-(y-6)^2+a^2=0 \\
& \Rightarrow(4-4 x+9-6 y)-(25-10 x)-(36-12 y)=0 \\
& \Rightarrow \quad 13-4 x-6 y-25+10 x-36+12 y=0 \\
& \Rightarrow \quad 6 x+6 y-48=0 \\
& \Rightarrow \quad x+y=\frac{48}{6}=8 \\
& \Rightarrow \quad x+y=8
\end{aligned}
$$
Since, the given circle cut orthogonally, therefore
$$
\begin{aligned}
& 2 g_1 g_2+2 f_1 f_2=C_1+C_2 \\
& \Rightarrow 2(-2)(-5)+2(-3)(-6)=\left((-2)^2+(-3)^2-a^2\right) \\
& +\left((-5)^2+(-6)^2-a^2\right) \\
& {\left[\because \text { radius }=\sqrt{g^2+f^2-c} \quad \therefore a^2=g^2+f^2-c \Rightarrow\right.} \\
& \left.c=g^2+f^2-a^2\right] \\
& \Rightarrow \quad 20+36=\left(13-a^2\right)+\left(61-a^2\right) \\
& \Rightarrow \quad 56=74-2 a^2 \\
& \Rightarrow \quad 2 a^2=74-56=18 \\
& \Rightarrow \quad a^2=9 \\
& \Rightarrow \quad a=3[\because \text { radius can't be negative }] \\
&
\end{aligned}
$$
Clearly, option (c) satisfy the Eq. (i)
Hence, the correct answer is option: 'C'.