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The point $(3,-4)$ lies on both the circles $x^2+y^2-2 x+8 y+13=0 \quad$ and $x^2+y^2-4 x+6 y+11=0$. Then, the angle between the circles is
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Verified Answer
The correct answer is:
$135^{\circ}$
Given circles are $x^2+y^2-2 x+8 y+13=0$
and $x^2+y^2-4 x+6 y+11=0$.
Here, $C_1=(1,-4), C_2=(2,-3)$,
$\begin{array}{lll}
\Rightarrow & & r_1=\sqrt{1+16-13}=2 \\
\text { and } & & r_2=\sqrt{4+9-11}=\sqrt{2}
\end{array}$
Now, $d=C_1 C_2=\sqrt{(2-1)^2+(-3+4)^2}=\sqrt{2}$
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2} \\
& =\frac{2-4-2}{2 \times 2 \times \sqrt{2}} \\
& =-\frac{1}{\sqrt{2}} \\
\Rightarrow \quad \theta & =135^{\circ}
\end{aligned}$
and $x^2+y^2-4 x+6 y+11=0$.
Here, $C_1=(1,-4), C_2=(2,-3)$,
$\begin{array}{lll}
\Rightarrow & & r_1=\sqrt{1+16-13}=2 \\
\text { and } & & r_2=\sqrt{4+9-11}=\sqrt{2}
\end{array}$
Now, $d=C_1 C_2=\sqrt{(2-1)^2+(-3+4)^2}=\sqrt{2}$
$\begin{aligned}
\therefore \quad \cos \theta & =\frac{d^2-r_1^2-r_2^2}{2 r_1 r_2} \\
& =\frac{2-4-2}{2 \times 2 \times \sqrt{2}} \\
& =-\frac{1}{\sqrt{2}} \\
\Rightarrow \quad \theta & =135^{\circ}
\end{aligned}$
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