In List-I, a pair of circles is given in A, B, C and in List-II, angle between those pair of circles is…
In List-I, a pair of circles is given in A, B, C and in List-II, angle between those pair of circles is given. Match the items from List-I to List-II.
$\begin{array}{lll}
\hline & { \text{List-I} } & \text{List-II} \\
\hline \text{A} & (x-2)^2+y^2=1 & \text{I} & 90^{\circ} \\
\hline \text{B} & x^2+y^2-6x-6y+9=0 & \text{II} & 135^{\circ} \\
& x^2+y^2-4x+4y-9=0 & & \\
\hline \text{C} & x^2+y^2+4x-14y+28=0 & \text{III} & 60^{\circ} \\
& x^2+y^2+4x-5=0 & & \\
\hline & & \text{IV} & 30^{\circ} \\
\hline
\end{array}$
The correct matching is
A B C
I II III
II I III
III I IV
III I IV
Solution
We know that, angle between two circles is given by
$\cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_1 r_2}$, where $r_1$ and $r_2$ are radius and $d$ is distance between centres.
$
\text { (A) } \begin{aligned}
\cos \theta & =\frac{(\sqrt{2})^2+(1)^2-\left[\sqrt{(2-2)^2+(1-0)^2}\right]^2}{2 \times \sqrt{2} \times 1} \\
& {\left[\therefore r_1=\sqrt{2}, r_2=1, c_1=(2,0), c_2=(2,1)\right] } \\
& =\frac{2+1-1}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}
\end{aligned}
$
$
\therefore \theta=45^{\circ} \text { or } 135^{\circ}
$
(B)
$
\begin{aligned}
\cos \theta= & \frac{(3)^2+(\sqrt{17})^2-\left[\sqrt{(3-2)^2+(3+2)^2}\right]^2}{2 \times 3 \times \sqrt{17}} \\
& {\left[\because r_1=3, r_2=\sqrt{17}, c_1=(3,3), c_2=(2,-2)\right] } \\
& =\frac{9+17-26}{6 \sqrt{17}}=0 \\
\theta & =90^{\circ}
\end{aligned}
$
(C)
$
\begin{aligned}
\cos \theta & =\frac{(5)^2+(3)^2-\left[\sqrt{(-2+2)^2+(7-0)^2}\right]^2}{2 \times 5 \times 3} \\
& {\left[\because r_1=5, r_2=3, c_1=(-2,7), c_2=(-2,0)\right] } \\
& =\frac{25+9-49}{30}=\frac{-15}{30}=\frac{-1}{2}
\end{aligned}
$
So, $\theta=120^{\circ}$ or $60^{\circ}$