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The-circle-x-2-y-2-2x-4y-20-0-is-inscribed-in-a-square-One-vertex-of-the-square-is-4-7-Find-the-coordinates-of-the-other-vertices-

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Question Number 174185 by nadovic last updated on 26/Jul/22
The circle x^2 +y^2 −2x−4y−20=0 is inscribed in a square. One vertex of the square is (−4, 7). Find the coordinates of the other vertices.
Thecirclex2+y2−2x−4y−20=0isinscribedinasquare.Onevertexofthesquareis(−4,7).Findthecoordinatesoftheothervertices.
Answered by Cesar1994 last updated on 26/Jul/22
x^2 +y^2 −2x−4y−20=0 ⇔ (x−1)^2 +(y−2)^2 =20+1+4 ⇔ (x−1)^2 +(y−2)^2 =25 the circle′s center is (1, 2) (((−3−7)/(6−(−4))))(((−3−7)/(−4−6)))=((−10)/(10))(((−10)/(−10)))=−1 the diametrically oposite vertex to A(−4,7) is B(1+5, 2−5)=B(6,−3) the others vertices are C(1+5, 2+5)=C(6,7) and D(1−5,2−5)=D(−4,−3) evaluating each point in (1) it is verified that they are on the circumference then prove that the diagonals AB and CD are perpendiculars one to the other (((−3−7)/(6−(−4))))(((−3−7)/(−4−6)))=((−10)/(10))(((−10)/(−10)))=−1 ∴ the ABCD is rhombus then side AC is horizontal and side BC is vertical side, so they are perpendicular ∴ ABCD is a square ■
x2+y2−2x−4y−20=0⇔(x−1)2+(y−2)2=20+1+4⇔(x−1)2+(y−2)2=25thecircle′scenteris(1,2)(−3−76−(−4))(−3−7−4−6)=−1010(−10−10)=−1thediametricallyopositevertextoA(−4,7)isB(1+5,2−5)=B(6,−3)theothersverticesareC(1+5,2+5)=C(6,7)andD(1−5,2−5)=D(−4,−3)evaluatingeachpointin(1)itisverifiedthattheyareonthecircumferencethenprovethatthediagonalsABandCDareperpendicularsonetotheother(−3−76−(−4))(−3−7−4−6)=−1010(−10−10)=−1∴theABCDisrhombusthensideACishorizontalandsideBCisverticalside,sotheyareperpendicular∴ABCDisasquare◼

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