Question

.The end points of diameter of a circle are (-3,1) and (5,7).The radius of the circle is

Answer 1

Given : The co – ordinates of end points of diameter of a circle are (-3,1 ) and (5,7) .

Need To Find : The Radius of Circle ?

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⠀☯︎⠀We've provided with the co – ordinates of end points of diameter of a circle ( i.e. (-3,1 ) and (5,7) ). We'll Find the Length of Diameter of Circle using Distance Formula .

⠀▪︎⠀We know that the Distance between two points ( x₁ , y₁ ) and ( x₂ , y₂ ) is – 

$\qquad \star\:\pmb{\underline {\boxed {\sf \: Distance \:=\: \sqrt{ \Bigg( x_2 - x_1 \Bigg)^2 \: +\:\Bigg( y_2 - y_1\:\Bigg)^2}\:\:}}}\:$

Where ,

• x₁ = – 3 ,
• x₂ = 5 ,
• y₁ = 1 &
• y₂ = 7 .

$\\\qquad \dag \underline {\frak{ Substituting \:Known \:Values \:in\:given \:Formula \:\::\:}}$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( x_2 - x_1 \bigg)^2 \: +\:\bigg( y_2 - y_1\:\bigg)^2}\:$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( 5 - ( - 3 ) \bigg)^2 \: +\:\bigg( 7 - 1 \:\bigg)^2}\:$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( 5 + 3 \bigg)^2 \: +\:\bigg( 7 - 1 \:\bigg)^2}\:$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( 8 \bigg)^2 \: +\:\bigg( 6 \:\bigg)^2}\:$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( 64 \: +\:36 \:\bigg)}\:$

$:\implies \:\sf \: Distance \:=\: \sqrt{ \bigg( 100 \:\bigg)}\:$

$:\implies \pmb{\:\sf \: Distance \:=\: 10\:units\:}\:$

Therefore,

⠀⠀⠀∴ Diameter of Circle is 10 units .

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$\qquad \bigstar \:\:\underline {\pmb{\sf Radius \:of\:Circle \:\::\:}}$

⠀⌬⠀ We know , that the Radius of the Circle is the half of the Diameter , i.e. Radius = Diameter/2 .

$\dashrightarrow \pmb{\sf Radius \:=\:\dfrac{Diameter}{2}\:}\\\\\\\dashrightarrow \sf Radius \:=\: \cancel{\dfrac{10}{2}}\:\\\\\\ \dashrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:Radius \:\:=\:5\:units\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf Hence ,\:Radius \:of\:the\:Circle \:is\:\pmb{\sf \:5\:units\:}\:.}$