Question

Find the angle of intersection of two circles?
$$x^2+y^2+2gx+2fy+c=0$$
$$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$$

Answer 1

Hint: To solve this question, we will, first of all, compute the centre and radius of both the circles using the formula $$c=({\displaystyle \frac{-g}{a}},{\displaystyle \frac{-f}{a}})$$ and $$r={\displaystyle \frac{1}{a}}\sqrt{g^2+f^2-c}$$ and then we will calculate the distance between $$c_1$$ and $$c_2$$ by using the distance between two points formula, $$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}.$$ Finally, we will use the formula of cos of the angle in a triangle to get the result.

Complete step-by-step solution:
We are given the two circles below.
$$x^2+y^2+2gx+2fy+c=0.....\left(i\right)$$
$$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0.....\left(ii\right)$$
They are given as

If the circles intersect at P, then the angle $$\theta$$ is the angle between the tangents to both the circles at the point P. $$c_1$$ and $$c_2$$ are the centres of the circles given by the equation (i) and (ii) respectively. And the standard equation of a circle is of the form
$$a{x}^2+b{y}^2+2gx+2fy+2hxy+c=0$$
where a, b, h, g, f and c are constants and the radius of the circle is given by
$$r={\displaystyle \frac{1}{a}}\sqrt{g^2+f^2-ca}$$
And the centre of the circle is given by
$$c=({\displaystyle \frac{-g}{a}},{\displaystyle \frac{-f}{a}})$$
Comparing this theory with our given equation (i) and equation (ii) of the circle, we have the centre as
$$c_1=(-g,-f)$$
$$c_2=(-g_1,-f_1)$$
And radius is given as,
$$r_1=\sqrt{g^2+f^2-c}$$
$$r_2=\sqrt{{g_1}^2+{f_1}^2-c_1}$$
Clearly, as visible by the above diagram, the distance ‘d’ between the circles is given by
$$d=\left|c_1{c}_2\right|$$
The formula of the distance between the two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is given by
$$d=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$$
Using this formula to find the distance $$d=\left|c_1{c}_2\right|$$ we have,
$$d=\sqrt{{(g-g_1)}^2+{(f-f_1)}^2}$$
$$\Rightarrow d=\sqrt{g^2+{g_1}^2-2g{g}_1+f^2+{f_1}^2-2f{f}_1}$$
$$\Rightarrow d=\sqrt{g^2+f^2+{g_1}^2+{f_1}^2+-2g{g}_1-2f{f}_1}$$
Now considering the triangle $$c_{1}P{c}_2$$ and the angle $$\alpha$$ between $$c_{1}P{c}_2.$$

Now, if triangle ABC is given as below, where the angle $$\theta$$ is $$\mathrm{\angle }BAC,$$ then if AB = c, AC = b and BC = a, then $$\cos \theta$$ is given by the formula $$\cos \theta ={\displaystyle \frac{b^2+c^2-a^2}{2bc}}.$$
Applying this formula on the angle $$\alpha$$ is given by
$$\cos \alpha ={\displaystyle \frac{{r_1}^2+{r_2}^2-d^2}{2r_1{r}_2}}$$
where $$\alpha$$ is the angle $$\mathrm{\angle }{c}_{1}P{c}_2.$$
Now, $$\theta$$ is the angle $$\mathrm{\angle }{A}^{{}^{\prime }}P{B}^{{}^{\prime }},$$ this is so as vertically opposite angles are equal and $$\mathrm{\angle }TPR=\theta .$$
$$\Rightarrow \mathrm{\angle }{A}^{{}^{\prime }}P{B}^{{}^{\prime }}=\mathrm{\angle }TPR=\theta \left[\text{vertically opposite angles}\right]$$
Now, because $$A^{{}^{\prime }}{c}_2$$ and $$B^{{}^{\prime }}{c}_1$$ are tangles to the circles of the centre $$c_1$$ and centre $$c_2$$ respectively, then $$\mathrm{\angle }{B}^{{}^{\prime }}P{c}_2=\mathrm{\angle }{A}^{{}^{\prime }}P{c}_1={90}^{\circ }.$$
Finally, we have a full circle angle that is $${360}^{\circ }.$$
$$\Rightarrow \mathrm{\angle }{B}^{{}^{\prime }}P{c}_2+\mathrm{\angle }{A}^{{}^{\prime }}P{c}_1+\mathrm{\angle }{A}^{{}^{\prime }}P{B}^{{}^{\prime }}+\mathrm{\angle }{c}_{1}P{c}_2={360}^{\circ }$$
$$\Rightarrow {90}^{\circ }+{90}^{\circ }+\alpha +\theta ={360}^{\circ }$$
$$\Rightarrow \alpha ={180}^{\circ }-\theta$$
Now, as $$\cos \alpha ={\displaystyle \frac{{r_1}^2+{r_2}^2-d^2}{2r_1{r}_2}}$$
$$\Rightarrow \cos ({180}^{\circ }-\theta )={\displaystyle \frac{{r_1}^2+{r_2}^2-d^2}{2r_1{r}_2}}$$
Hence, the angle between the circle is given by $$\cos ({180}^{\circ }-\theta )={\displaystyle \frac{{r_1}^2+{r_2}^2-d^2}{2r_1{r}_2}}$$ where $$r_1$$ is the radius of the circle $$x^2+y^2+2gx+2fy+c=0$$ and $$r_2$$ is the radius of the circle having equation $$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$$ and d is the distance between the centre $$c_1$$ and $$c_2$$ of both the circles.

Note: Because $$\alpha$$ is the angle between the centre $$c_1$$ of the circle $$x^2+y^2+2gx+2fy+c=0$$ and the centre $$c_2$$ of the circle $$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0.$$ Therefore, we needed to calculate the value of $$\alpha .$$ This $$\alpha$$ will give the angle between two given circles. Also, in such cases always the angle from the centre of the two circles is measured.