Q.Show that the angle between the circles $x^2 + y^2 = a^2$, $x^2 + y^2 = ax + ay$ is $\frac{3\pi}{4}$.

Using $\cos\theta=\dfrac{r_1^2+r_2^2-d^2}{2r_1r_2}$, the angle between $x^2+y^2=a^2$ and $x^2+y^2=ax+ay$ comes out to $\dfrac{\pi}{4}$, confirmed independently by directly finding the tangent lines at a point of intersection.

Setting up. Circle 1: $x^2+y^2-a^2=0$, centre $O_1=(0,0)$, $r_1=a$.

Circle 2: $x^2+y^2-ax-ay=0$, i.e. $\left(x-\tfrac a2\right)^2+\left(y-\tfrac a2\right)^2=\tfrac{a^2}{2}$, centre $O_2=\left(\tfrac a2,\tfrac a2\right)$, $r_2=\tfrac{a}{\sqrt2}$.

Distance between centres: $d=\sqrt{\left(\tfrac a2\right)^2+\left(\tfrac a2\right)^2}=\dfrac{a}{\sqrt2}=r_2$.

Angle formula.

$\cos\theta = \dfrac{r_1^2+r_2^2-d^2}{2r_1r_2} = \dfrac{a^2+\tfrac{a^2}{2}-\tfrac{a^2}{2}}{2\cdot a\cdot \tfrac{a}{\sqrt2}} = \dfrac{a^2}{\tfrac{2a^2}{\sqrt2}} = \dfrac{\sqrt2}{2}=\dfrac{1}{\sqrt2}$

So $\theta=\dfrac{\pi}{4}$.

Independent check (direct geometry). Substituting $x^2+y^2=a^2$ into circle 2's equation gives the common chord $x+y=a$; solving with the circle gives intersection points $(a,0)$ and $(0,a)$. At $P=(a,0)$:

• Tangent to circle 1 (perpendicular to radius $O_1P$, which lies along the $x$-axis) is the vertical line $x=a$.
• Implicit differentiation of circle 2 gives slope $\dfrac{dy}{dx}=\dfrac{a-2x}{2y-a}$; at $(a,0)$ this is $\dfrac{-a}{-a}=1$, so the tangent to circle 2 has slope $1$ (a $45^\circ$ line).

The angle between a vertical line and a line of slope $1$ is $45^\circ=\dfrac{\pi}{4}$ — matching the formula exactly.

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