If the angle between the circles $x^2+y^2-2 x-4 y+c=0$ and $x^2+y^2-4 x-2 y+4=0$ is $60^{\circ}$, then $c$…
If the angle between the circles $x^2+y^2-2 x-4 y+c=0$ and $x^2+y^2-4 x-2 y+4=0$ is $60^{\circ}$, then $c$ is equal to
- $\frac{3 \pm \sqrt{5}}{2}$
- $\frac{6 \pm \sqrt{5}}{2}$
- $\frac{9 \pm \sqrt{5}}{2}$
- $\frac{7 \pm \sqrt{5}}{2}$
Solution
We have,
$\begin{aligned}
& \mathrm{C}_1: x^2+y^2-2 x-4 y+c=0 ... (i)\\
& \mathrm{C}_2: x^2+y^2-4 x-2 y+4=0... (ii)
\end{aligned}$
So, centre of circle $\mathrm{C}_1=(1,2), r_1=\sqrt{5-c}$ and centre of circle $\mathrm{C}_2=(2,1), r_2=1$
Now, angle between the two circles,
$\begin{aligned}
\cos \theta & =\frac{\left(\mathrm{C}_1 \mathrm{C}_2\right)^2-\left(r_1^2+r_2^2\right)}{2 r_1 r_2} \\
\cos \theta & =\frac{\left[(2-1)^2+(1-2)^2\right]-(5-c+1)}{2 \sqrt{5}-c(1)}
\end{aligned}$
Given, $\theta=60^{\circ}$
$\begin{aligned}
& \cos 60^{\circ}=\frac{1+1-5+c-1}{2 \sqrt{5-c}} \\
& \frac{1}{2}=\frac{c-4}{2 \sqrt{5-c}} \\
& \sqrt{5-c}=c-4
\end{aligned}$
On squaring both sides, we get
$\begin{aligned}
& 5-c=c^2-8 c+16 \\
& c^2-5 c+11=0 \\
& c=\frac{7 \pm \sqrt{49-44}}{2} \\
& c=\frac{7 \pm \sqrt{5}}{2}
\end{aligned}$
Asked in: AP EAMCET 2016
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