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Question Number 183107 by mr W last updated on 21/Dec/22

Commented by mr W last updated on 20/Dec/22

$$findtheequationofthecirclewith$$$$radiusrandcenteratC(a,b,c).the$$$$normaloftheplaneinwhichthe$$$$circleliesis\overrightarrow{n}=(\alpha ,\beta ,\gamma ).$$

Answered by mr W last updated on 21/Dec/22

Commented by mr W last updated on 21/Dec/22

$$theeqn.oftheplanecontainingthe$$$$circleis$$$$\alpha (x-a)+\beta (y-b)+\gamma (z-c)=0$$$$withz=c:$$$$\alpha (x-a)+\beta (y-b)+\gamma (c-c)=0$$$$\alpha x+\beta y=\alpha a+\beta b$$$$\overrightarrow{u}=(-\beta ,\alpha ,0)$$$${\overrightarrow{u}}_1=(-\frac{\beta }{\sqrt{{\alpha }^2+{\beta }^2}},\frac{\alpha }{\sqrt{{\alpha }^2+{\beta }^2}},0)$$$$\overrightarrow{v}=\overrightarrow{n}\times \overrightarrow{u}=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ -\beta & \alpha & 0\end{array}\right]=(-\alpha \gamma ,\beta \gamma ,{\alpha }^2+{\beta }^2)$$$${\overrightarrow{v}}_1=(-\frac{\alpha \gamma }{\sqrt{({\alpha }^2+{\beta }^2)({\alpha }^2+{\beta }^2+{\gamma }^2)}},\frac{\beta \gamma }{\sqrt{({\alpha }^2+{\beta }^2)({\alpha }^2+{\beta }^2+{\gamma }^2)}},\frac{\sqrt{{\alpha }^2+{\beta }^2}}{\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}})$$$$eqn.ofcirclewithparameter\theta :$$$$P(x,y,z)=(a,b,c)+r\cos \theta {\overrightarrow{u}}_1+r\sin \theta {\overrightarrow{v}}_1$$$$x=a-\frac{\beta r\cos \theta }{\sqrt{{\alpha }^2+{\beta }^2}}-\frac{\alpha \gamma r\sin \theta }{\sqrt{({\alpha }^2+{\beta }^2)({\alpha }^2+{\beta }^2+{\gamma }^2)}}$$$$y=b+\frac{\alpha r\cos \theta }{\sqrt{{\alpha }^2+{\beta }^2}}+\frac{\beta \gamma r\sin \theta }{\sqrt{({\alpha }^2+{\beta }^2)({\alpha }^2+{\beta }^2+{\gamma }^2)}}$$$$z=c+\frac{\sqrt{{\alpha }^2+{\beta }^2}r\sin \theta }{\sqrt{{\alpha }^2+{\beta }^2+{\gamma }^2}}$$

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