Cardiode $r=a(1-cosɵ)-------(1)$

Diff. w.r.t. ɵ

$\therefore \dfrac {dr}{d\theta}=a(0+\sin\theta)\\ =a\sin \theta$

Circle : $r=a\cos\theta---- (2)$

Converting polar form to Cartesian form put

$\cos\theta=\dfrac xr\space and \space x^2+y^2=r^2$

$r=\dfrac {ax}r\\ r^2=ax\\ \therefore x^2+y^2-ax=0\\ Here\space 2g=-a ,2f=0,c=0\\ \therefore center =(-g,-f)=\Bigg(\dfrac a2,0\Bigg)\\ radius=\sqrt{g^2+f^2-c}\\ =\sqrt{\dfrac {a^2}{2^2}=\dfrac a2}$

From (1) and (2)

$$a(1-\cos θ) = a\cos θ$$

$\therefore 1=2\cos\theta \\ \cos \theta=1/2 \\ \theta =\dfrac \pi3 \space or \space \theta=\dfrac {-\pi}3$

Cardiode intersects the circle.

At $\theta=\pm \dfrac \pi3$

Now length of arc of cardiode outside the circle

= 2 × length of arc of cardiode outside circle and above x-axis

$$=2\int\limits_{\theta_1}^{\theta_2}\sqrt{r^2+\Bigg(\dfrac {dr}{d\theta}\Bigg)^2} d\theta$$

$=2\int\limits_{\frac \pi3}^{\pi}\sqrt{a^2(1-\cos\theta)^2+a^2\sin^2\theta }\space d\theta\\ =2\int\limits_{\frac \pi3}^{\pi} a\sqrt{1-2\cos \theta + \cos^2 \theta + \sin^2\theta }\space d\theta \\ =2\int\limits_{\frac \pi3}^{\pi}a\sqrt{2-2\cos\theta }\space d\theta \\ =2\sqrt 2 a \int\limits_{\frac \pi3}^{\pi} \sqrt{2\sin^2\dfrac {\theta}2}\space d\theta \\ =4a \int\limits_{\frac \pi3}^{\pi}\sin \dfrac {\theta} 2.d\theta \\ =4a\Bigg[\dfrac {-\cos\dfrac {\theta}2}{\frac 12}\Bigg]_{\frac \pi3}^{\pi} \\ =8a\Big\{\Big[-\cos \dfrac \pi2\Big]-\Big[-\cos\dfrac \pi6\Big]\Big\}\\ =8a\Bigg[0+\dfrac {\sqrt3}2\Bigg]\\ =4a\sqrt 3$