Question

If the angle between the circles $x^2+y^2-2x-4y+c=0$ and $x^2+y^2-4x-2y+4=0$ is $60^{\circ }$, then $c$ is equal to

• A. $\frac{3\pm \sqrt{5}}{2}$
• B. $\frac{6\pm \sqrt{5}}{2}$
• C. $\frac{9\pm \sqrt{5}}{2}$
• D. $\frac{7\pm \sqrt{5}}{2}$

Answer 1

Answer: (D)

Given,
$x^2+y^2-2x-4y+c=0...(i)$
and $x^2+y^2-4x-2y+4=0...(ii)$
Centre of first circle is $(1,2)$ and radius of this circle is $\sqrt{1+4-c}=\sqrt{5-c}$
Centre of second circle is $(2,1)$ and radius of this circle is$\sqrt{4+1-4}=\sqrt{1}=1$
Hence, $C_1\equiv (1,2),r_1=\sqrt{5-c}$
and $C_2=(2,1),r_2=1$
Angle between the two circles is
$\cos \theta =\frac{{\left(C_1{C}_2\right)}^2-\left(r_1^2+r_2^2\right)}{2r_1{r}_2}$
$\cos \theta =\frac{\left[(2-1{)}^2+(1-2{)}^2\right]-(5-c+1)}{2\sqrt{5-c}(1)}$
$\Rightarrow \cos 60^{\circ }=\frac{1+1-5+c-1}{2\sqrt{5-c}}$
$\Rightarrow \frac{1}{2}=\frac{c-4}{2\sqrt{5-c}}$
$\Rightarrow \sqrt{5-c}=c-4$
On squaring both sides, we get
$5-c=c^2-8c+16$
$\Rightarrow c^2-7c+11=0$
$\therefore c=\frac{7\pm \sqrt{49-44}}{2}$
$=\frac{7\pm \sqrt{5}}{2}$