Show that the curves are circle.

Question

For example, this question is also similar to the previous question I have asked, which is the following link; How to show the curves are conics.

Question:

Solve the equation $$\frac{dx}{cy-bz}=\frac{dy}{az-cx}=\frac{dz}{bx-ay}$$

And show that they are circles.

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I solve the question

Firstly I found the integral curves.

Let $$P=cy-bz$$ $$Q=az-cx$$ $$R=bx-ay$$

Let's take $P'=a$ $Q'=b$ and $R'=c$ such that $PP'+QQ'+RR'=0$.

Then $c_1=ax+by+cz$

Let $$\frac{xdx+ydy}{cxy-bxz+ayz-cxy}=\frac{dz}{bx-ay}$$ $$\frac{xdx+ydy}{z(ay-bx)}=\frac{dz}{-(ay-bx)}$$

Then $c_2=x^2+y^2+z^2$

However, honestly I dont know how to show that they are circles.

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And I cannot show the integral curves at previous question which I posted its link here above.

Please help me my questions. I am new learner of PDE.

Answer 1

Hint: You have two equations; one of a plane and one of a sphere. They will always intersect at a circle.

Answer 2

As @Harry suggested: The intersection of the plane and the sphere is the quadratic equation $$(c^2-a^2)x^2+(c^2-b^2)y^2+2abxy-2c_1(ax+by)=c^2c_2-c_1^2$$ and this is an ellipse (circle is its degenerate form) provided that $\Delta=(ab)^2-(c^2-a^2)(c^2-b^2)=c^2[a^2+b^2-c^2]>0$. This holds true when $c<\sqrt{a^2+b^2}$.