Question

Find the angle between the circles
$S:{{x}^{2}}+{{y}^{2}}-4x+6y+11=0$ and $S':{{x}^{2}}+{{y}^{2}}-2x+8y+13=0$

Answer 1

Hint:First compare the given equations of circle with the general equation of the circle, that is, ${{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}={{r}^{2}}$, to find out the centre and radius of both the circle. Then apply the formula $\cos \theta =|\dfrac{{{d}^{2}}-{{r}_{1}}^{2}-{{r}_{2}}^{2}}{2{{r}_{1}}{{r}_{2}}}|$.

Complete step-by-step answer:
The given equations of circle are:
$S:{{x}^{2}}+{{y}^{2}}-4x+6y+11=0..................(i)$
$S':{{x}^{2}}+{{y}^{2}}-2x+8y+13=0...............................(ii)$
From equation (i) & (ii), we will find their centres ${{C}_{1}}$ and ${{C}_{2}}$ first, respectively for circles S and S’.
Adding $4+9$ to both sides of the equation, we get :
Equation (i): ${{x}^{2}}-4x+4+{{y}^{2}}+6y+9+11=4+9$
\[\begin{align}
& {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=-11+4+9 \\
& \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}=2 \\
\end{align}\]
By comparing this equation to the form ${{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}={{r}^{2}}$, where the centre of the circle is $({{x}_{0}},{{y}_{0}})$, and its radius is $r$, we get ${{C}_{1}}\left( 2,-3 \right)$.
Now, for equation (ii): ${{x}^{2}}+{{y}^{2}}-2x+8y+13=0$
Let’s add $1+16$ to both sides. Doing so, we get :
$\left( {{x}^{2}}-2x+1 \right)+{{y}^{2}}+8y+16+13=16+1$
\[\begin{align}
& {{\left( x-\underline{1} \right)}^{2}}+{{\left( y+\underline{4} \right)}^{2}}=-13+16+1 \\
& \Rightarrow {{\left( x-\underline{1} \right)}^{2}}+{{\left( y+\underline{4} \right)}^{2}}=4 \\
\end{align}\]
Hence we get ${{C}_{2}}(1,-4)$
Now, we will find the radius; we get
\[{{r}_{1}}=\sqrt{g_{1}^{2}+f_{1}^{2}-C}\] and ${{r}_{2}}=\sqrt{{{g}_{2}}^{2}+{{f}_{2}}^{2}-C}$
As given with centre ${{C}_{1}}\left( 2,-3 \right)$ and ${{C}_{2}}\left( 1,-4 \right)$
Of the two circle equation in standard form, the standard equation is actually given by \[\]
So, we know that the Standard equation of a circle is
Now, for radius of ‘S’ for which the centre ${{C}_{1}}(2,-3)$ is given by,
\[\begin{align}
& {{r}_{1}}=\sqrt{{{g}_{1}}^{2}+{{f}_{1}}^{2}-C} \\
& =\sqrt{{{\left( 2 \right)}^{2}}+{{\left( 3 \right)}^{2}}-11} \\
\end{align}\]
$=\sqrt{2}$ units
Now, for radius of S’ for ${{C}_{2}}(1,-4)$ is given by;
\[\begin{align}
& {{r}_{2}}=\sqrt{{{g}_{2}}^{2}+{{f}_{2}}^{2}-C} \\
& =\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -4 \right)}^{2}}-13} \\
\end{align}\]
$\sqrt{4}=2$ units
Now, we find the distance between the centres using distance formula, that says that the distance between two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is given by :
$d=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Applying this formula for ${{C}_{1}}$ and ${{C}_{2}}$, we’ll see that :
$d=\sqrt{{{\left( 2-1 \right)}^{2}}+{{\left( -3+4 \right)}^{2}}}$
$\begin{align}
& =\sqrt{1+1} \\
& =\sqrt{2} \\
\end{align}$
Now, for finding the angle, we will use the formula $\to \cos \theta =\left| \dfrac{{{d}^{2}}-{{r}_{1}}^{2}-{{r}_{2}}^{2}}{2{{r}_{1}}{{r}_{2}}} \right|$, where $\theta $ is the angle between the circles. So, applying that formula here, we get:
\[\begin{align}
& \cos \theta =\left| \dfrac{{{\left( \sqrt{2} \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}-{{\left( 2 \right)}^{2}}}{2\times \sqrt{2}\times 2} \right| \\
& =\left| \dfrac{2-2-4}{4\sqrt{2}} \right| \\
& =\left| \dfrac{-{4}}{{4}\sqrt{2}} \right| \\
& \cos \theta =\pm \dfrac{1}{\sqrt{2}} \\
& \theta ={{\cos }^{-1}}\left( \pm \dfrac{1}{\sqrt{2}} \right) \\
\end{align}\]
\[=45{}^\circ ,135{}^\circ \]
We won’t consider the angles that are greater than $180{}^\circ $ over here, since the angle between two circles can at max be $180{}^\circ $ only.
$\therefore $ Therefore, the angle between two circle is $45{}^\circ or135{}^\circ $

Note: Students should remember the formula i.e $\cos \theta =\left| \dfrac{{{d}^{2}}-{{r}_{1}}^{2}-{{r}_{2}}^{2}}{2.{{r}_{1}}.{{r}_{2}}} \right|$ to solve these types of questions.The general form of the equation of circle is given as ${x^2+y^2+2gx+2fy+c}$.The centre of circle is $(-g,-f)$ and radius of circle is ${\sqrt{g^2+f^2-c}}$.We can directly calculate centre from the equation $S:{{x}^{2}}+{{y}^{2}}-4x+6y+11=0$ equating $-4=2g$ and $6=2f$ we get $g=-2$ and $f=3$ so centre of circle is $(-g,-f)$ i.e $(2,-3)$.