This question was briefly on MSE yesterday, asked by someone else.

If we have two circles $$(x-x(A))^2+(y-y(A))^2=r_1^2$$ $$(x-x(B))^2+(y-y(B))^2=r_2^2$$ tangent at $C$ with the common tangent line going through the origin then the other tangents from the origin have segments from the origin to the tangent points $D, E$ equal and equal to $\sqrt{x(A)^2+y(A)^2-r_1^2}=\sqrt{x(B)^2+y(B)^2-r_2^2},$ the square root of the constant term in the expanded equations.


Now if the other tangents are perpendicular


then the question is to prove


My attempt: I have verified it using geogebra for a varying line and varying point $A$ on the bisector line, the rest (B on the other bisector line and the perpendicular line from A through the circle tangent point) being determined by the constraints. The strategy has been to look for areas, hoping two would be the same and give the identity, finding


Now this is $|x(A)y(B)-y(A)x(B)|,$ and using the auxiliary circle through the origin, $A,B,C,D,E;$ $|x(A)|=|y(B)|$ and $|x(A)x(B)+y(A)y(B)|=|\pm |y(B)|x(B)+y(A)y(B)|=|y(B)||\pm |x(B)|\pm |y(A)||.$ It seems I need a new idea to get over the finish line.


From QEDs answer I see that what needs to be proven is $|x(A)y(B)-y(A)x(B)|=2\cdot \operatorname{Area}(\triangle{OAB}),$ but I don't see why I can take $B$ to be the midpoint on the square of $|y(B)|.$

Anyway, WLOG I can scale to axis-aligned square of side $1$ cut away its upper right corner: square $\operatorname{Area}(\triangle{OAB})=1-(\frac12 (1-x(B))(1-y(A))+\frac12 x(B)+\frac12 y(A))$ so $2\cdot \operatorname{Area}(\triangle{OAB})=1-x(B)y(A)$ and we're done.

  • $\begingroup$ $B$ is the midpoint of the square because because the side on which $B$ lies is the diameter of the circle. $\endgroup$
    – QED
    Sep 25 '20 at 18:01

Let us call the radius of the circle centered at $A$ as $r_A=|y(A)|$ and the radius of the circle centered at $B$ be $r_B=|x(B)|$. Let the length of the common tangent be $T=|x(A)|=|y(B)|$. Then we see that $|x(B)y(B)+x(A)y(A)|=T.(r_A+r_B)=2.\mathrm{area}(OAB)$. Your problem is basically reduced to this:

For a square $ABCD$, let $E$ be the midpoint of $CD$ and $F$ be a point on $BC$. Also $EI\parallel AD$, $FG\parallel AB$, and $EI$ and $FG$ intersect at $H$. Then prove $$\mathrm{area}(ABCD)-\mathrm{area}(AIHG)=2.\mathrm{area}(AFE)$$ enter image description here

This is a much simpler problem and quite easy to handle. Can you solve it?


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