  Question: Area of circle touching the line |x-2|+|y-3|=4 is

Posted by: rohan on 26.07.2021 Harendran from Coimbatore
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Yuvaraj K. from Pune
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Satheesh O. from Kochi
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From the figure hypotenuse is 8 units

so by pythagoras theorem

a^2+a^2=8^2

2a^2=64

a^2=32

a=4√2

r=1/2 a=2√2

S
Satheesh O. from Kochi
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figure is S
Satheesh O. from Kochi
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Equation of the lines |x-2|+|y-3|=4

First find equation of the circle touching the lines |x-2|+|y-3|=4

First, draw the lines

First, draw the lines.

So from the figure, it's clear that the centre of these lines is (2, 3)

Thus the circle should have the same centre as of these lies because the circle touches these lines.

The standard equation of a circle is given by,

(x - h)² + (y - k)² = r²

The radius of the circle will be equal to half of the side of the square formed by these lines.

So, we have,

(x - 2)² + (y - 3)² = 8

Therefore, the equation of the circle touching the lines |x-2|+|y-3|=4 is (x - 2)² + (y - 3)² =(2√2)^2

Therefore area of the circle =pi*r^2=3.14*8=25.12 sq.unit

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