Show that the angle between the circles
$x^{2} + y^{2} = a^{2}, x^{2} + y^{2} = a x + a y$ is $\frac{3 π}{4}$

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Solution

$S_{1} \to x^{2} + y^{2} = a^{2}$ ... $(i)$
$S_{2} \to x^{2} + y^{2} - a x - a y = 0$ ... $(i i)$
Form equation $(i)$
$y = \sqrt{a^{2} - x^{2}}$
Hence
$x^{2} + (a^{2} - x^{2}) - a x - a (\sqrt{a^{2} - x^{2}}) = 0$
$a^{2} - a x - a (\sqrt{a^{2} - x^{2}}) = 0$
$a - x - (\sqrt{a^{2} - x^{2}}) = 0$
$(a - x) = \sqrt{a^{2} - x^{2}} = 0$
$(a - x) - \sqrt{(a - x) (a + x)} = 0$
$\sqrt{a - x} (\sqrt{a - x} - \sqrt{a + x}) = 0$
$a = x$ or $(\sqrt{a - x} - (\sqrt{a + x}) = 0$
$a - x = a + x$ or $x = 0$
Therefore, $x_{1} = a$ and $x_{2} = 0$ . Similarly $y_{1} = 0$ and $y_{2} = a$ .
Thus the points of intersection as
$(a, 0)$ and $(0, a)$ .
Now the slope of the tangent of $S_{1}$ at $(0, a)$
$2 x + 2 y y^{'} = 0$
$y^{'} = \frac{- x}{y}$
$y_{x = (0, a)}^{'} = 0 = m_{1}$ .
Now the slope of the tangent of $S_{2}$ at $(0, a)$
$2 x + 2 y y^{'} - a - a y^{'} = 0$
$y^{'} (2 y - a) = a - 2 x$
$y^{'} = \frac{- (2 x - a)}{(2 y - a)}$
$y_{x = (0, a)}^{'} = 1 = m_{2}$ .
Hence angle between the circles at $(0, a)$ is
$tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} . m_{2}}$
$= \frac{0 - 1}{1 + (0 \times 1)}$
$= - 1$ .
$tan θ = - 1$ or $θ = \frac{3 π}{4}$ .

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