On Smarandache’s Sphere Edited by Prof. Mihaly Bencze Áprily Lajos College Brașov, Romania Through one of the intersecting points of two circles we draw a line that intersects a second time the circles in the points M 1 and M 2 respectively. Then the geometric locus of the point M which divides the segment M 1M 2 in a ratio k (i.e. M1M = k⋅MM2) is the circle of center O (where O is the point that divides the segment of line that connects the two circle centers O1 and respectively O2 into the ratio k, i.e. O1O = k ⋅ OO2 ) and radius OA, without the points A and B. Proof Let O1 E ⊥ M 1M 2 and O2 F ⊥ M 1 M 2 . Let O ∈ O1O2 such that O1O = k ⋅ OO2 and M ∈ M 1M 2 , where M 1M = k ⋅ MM 2 . A
G F
M2
E M1
O1
O O2 B Fig. 1.
We construct OG ⊥ M 1M 2 and we make the notations: M 1 E ≡ EA = x and AF ≡ FM 2 = y . Then, AG ≡ GM , because − x + ky k AG = EG − EA = ( x + y) − x = k +1 k +1 and − x + ky − x + ky k . = GM = M 1 M − M 1 A − AG = ( 2x + 2 y ) − 2x − k +1 k +1 k +1 Therefore we also have OM ≡ OA .