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Question Number 199391 by Calculusboy last updated on 03/Nov/23

Answered by deleteduser1 last updated on 03/Nov/23

(x−5)^2 +(y−5)^2 =10⇒(x−5)^2 +(5−2x)^2 =10  ⇒5(x^2 −6x+8)=0⇒x=4 or 2⇒(x,y)=(4,2);(2,6)    (5,5);(4,2)⇒slope of tangent line passing through  (4,2)=((−1)/3)⇒((y−2)/(x−4))=((−1)/3)⇒3y−6=−x+4⇒3y+x=10  (5,5);(2,6)⇒slope of tangent line passsing (2,6)  =3⇒((y−6)/(x−2))⇒y=3x  x=1;y=3 at the point of intersection of the two  tangents

(x5)2+(y5)2=10(x5)2+(52x)2=105(x26x+8)=0x=4or2(x,y)=(4,2);(2,6)(5,5);(4,2)slopeoftangentlinepassingthrough(4,2)=13y2x4=133y6=x+43y+x=10(5,5);(2,6)slopeoftangentlinepasssing(2,6)=3y6x2y=3xx=1;y=3atthepointofintersectionofthetwotangents

Commented by Calculusboy last updated on 03/Nov/23

thanks sir

thankssir

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