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How can I find the point on the circumference of the circle (A) that is intersected by the line from the center through a point inside the circle. Only the radius, center, and position of the point inside the circle is provided, not the angle.

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    $\begingroup$ Point $A$ is the intersection of the circle with the line $y=\dfrac{25}{60}x$ $\endgroup$
    – Qurultay
    Mar 30, 2020 at 11:29

2 Answers 2

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Let $O(0,0)$ the centre of the circle. The equation of the circle is: $$x^2+y^2=150^2$$ And the equation of the line $OA$ is: $$y=\frac{60}{25}x=\frac{12}{5}x$$ To find the coord of the point $A$, you have to solve this system of two equations in two unknows: $$\left\{\begin{matrix} x^2+y^2=150^2 \\ y=\frac{12}{5}x \end{matrix}\right.$$ There are two different solutions: $A_0\left(-\frac{750}{13},-\frac{1800}{13}\right) \lor A_1\left(\frac{750}{13},\frac{1800}{13}\right)$.

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For any point $p(p_1,p_2)\neq(0,0)$ (not necessarily inside the circle) multiply the unit vector in direction $p$ with circle's radius to obtain $$\frac{(p_1,p_2)}{\sqrt{p_1^2+p_2^2}}\cdot150.$$

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