Q93 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics
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 & \multicolumn{1}{c}{ List-I } & List-II \\
\hline A (x-2)^2+y^2=2(x-2)^2+(y-1)^2=1 & I & 90^{\circ} \\
\hline B x^2+y^2-6 x-6 y+9=0 & II 135^{\circ} \\
x^2+y^2-4 x+4 y-9=0 & & \\
\hline C x^2+y^2+4 x-14 y+28=0 & III 60^{\circ} \\
x^2+y^2+4 x-5=0 & & \\
\hline & & IV 30^{\circ} \\
\hline
\end{array}
The correct matching is
A B C
- AI II III
- Correct answerBII I III
- CIII I IV
- DIII 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 $60^{\circ}$.
Hence, the correct answer is option: 'B'.