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Question

The equation of the circle with center (2,1) and touching the line 3x+4y=5 is:


A

x2+y2-4x-2y+5=0

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B

x2+y2-4x-2y-5=0

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C

x2+y2-4x-2y+4=0

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D

x2+y2-4x-2y-4=0

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Solution

The correct option is C

x2+y2-4x-2y+4=0


Explanation for the correct answer :

Given the centre of the circle is (2,1) and the circle touches the line 3x+4y=5.

The distance between the centre and this line will be the radius.

Calculating the distance between (2,1) and 3x+4y=5:

The distance formula is

r=ax+by+ca2+b2

Substituting the values we get,

r=3x+4y-532+42a=3,b=4,c=5fromtheequationr=3(2)+4(1)-525givenx=2,y=1r=55r=1

Hence, the equation of the circle is x-h2+y-k2=a2

(x-2)2+(y-1)2=rh=2,k=1x2+y2-4x-2y+5=1x2+y2-4x-2y+4=0

Therefore, the correct answer is Option (C).


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