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Question: Radius of the circle \(( x - 1 ) ( x - 3 ) + ( y - 2 ) ( y - 4 )\) \(= 0\) is....

Radius of the circle (x−1)(x−3)+(y−2)(y−4)( x - 1 ) ( x - 3 ) + ( y - 2 ) ( y - 4 ) =0= 0 is.

A

2

B

2\sqrt { 2 }

C

3

D

222 \sqrt { 2 }

Answer

2\sqrt { 2 }

Explanation

Solution

Equation of the circle is (x−1)(x−3)+(y−2)(y−4)=0( x - 1 ) ( x - 3 ) + ( y - 2 ) ( y - 4 ) = 0

So, radius =12(3−1)2+(4−2)2=128=222=2= \frac { 1 } { 2 } \sqrt { ( 3 - 1 ) ^ { 2 } + ( 4 - 2 ) ^ { 2 } } = \frac { 1 } { 2 } \sqrt { 8 } = \frac { 2 \sqrt { 2 } } { 2 } = \sqrt { 2 }.