Question

find the equation of the circle which has A(1,3) and B(4,5) as opposite ends of a diameter? Find also the equation of the perpendicular diameter?
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Answer 1

HERE IS YOUR ANSWER FRIEND:

Diameter of circle is AB.

Diameter of circle is AB.Let the center of circle be O.

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+c

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-5/2=−3/2∗4+c

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-5/2=−3/2∗4+cc=17/2

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-5/2=−3/2∗4+cc=17/2perpendicular diameter:-

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-5/2=−3/2∗4+cc=17/2perpendicular diameter:-y=2−3x+217

Diameter of circle is AB.Let the center of circle be O.A(1,3) B(4,5)Slope of AB:-4−15−3=32M.P of AB=0=(5/2,4)We need the equation of diameter COD⊥ AOB⇒ Slope of COD:- −3/2COD passes through O. Eg. of COD:-y=−3/2x+cO lies on COD:-5/2=−3/2∗4+cc=17/2perpendicular diameter:-y=2−3x+2172y+3x−17=0.²

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