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If this can easily be derived from an already answered question please be so kind to point me there. I tried but couldn't find an equivalent problem.

I have one circle $D$ with the origin as center fully defined by its radius $r_D$. I have a given point $P$ which is outside $D$.

Now I want to construct a circle $H$ with known radius $r_H > r_D$, such that $P$ is on the circle and $D$ is enclosed inside $H$ and $H$ is touching $D$.

Where is the circle center of $H$/how do I calculate the point where $D=H$?

There are two solutions (though I just need one).

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  • $\begingroup$ connect center of $D$ (let's say point $O$) with $P$. Now draw a perpendicular to $OP$ that goes thru $O$. The perpendicular $OA$ will intersect $D$ at two points $A$ and $B$ (hence two solutions). Triange $OPA$ is a right triangle. Find $\angle{OPA}$. Construct angle equal to $90-2 \cdot OPA$ that has $P$ as vertex and $OP$ as one of the sides (again, there are two ways to construct the angle, choose wisely), The other side of the angle will intersect line $OA$ at some point, that point will be the center of $H$. $\endgroup$
    – Vasili
    Jul 30, 2018 at 13:35
  • $\begingroup$ Hi, Thanks for your suggestion. The circle center of $H$ cannot be on the line $OA$ though. Your suggestion sets $r_H=3*r_D$ (if I'm not mistaken) but the radius $r_H$ is actually known and fixed. $\endgroup$
    – the_toast
    Jul 30, 2018 at 14:11

1 Answer 1

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Notice that $H$ must touch (i.e. not intersect) $D$. This means it's center is located on a circle around the center of $D = (0,0)$ with the radius $r_H - r_D$. Simultaniously it must be on a circle with radius $r_D$ with center $P$. Now your problem is reduced to finding the intersection points of two circles.

Say: $(c_x,c_y)$ is the center of $H$ and wlog $P = (P_x,0)$. You have the equations: $$ c_x^2+c_y^2 = (r_D - r_H)^2 \text{ and } (c_x - P_x)^2+c_y^2 = r_H^2 $$ The rest should be simple, i guess.

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  • $\begingroup$ This is a beautifully simple solution that none of the 5 people I asked came up with. Props! $\endgroup$
    – the_toast
    Jul 30, 2018 at 14:17
  • $\begingroup$ Glad i could help! $\endgroup$
    – denklo
    Jul 30, 2018 at 14:28

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