Equation of circle touching the line |x-2|+|y-3|=4 will be Options: A . &nbsp ( x − 2 ) 2 + ( y − 3 ) 2 = 12 B . &nbsp ( x − 2 ) 2 + ( y − 3 ) 2 = 4 C . &nbsp ( x − 2 ) 2 + ( y − 3 ) 2 = 10 D . &nbsp ( x − 2 ) 2 + ( y − 3 ) 2 = 8

Answer: Option D : D Perpendicular distance from centre to tangent = radius r = | 2 + 3 − 9 | √ 2 = 4 √ 2 = 4 √ 2 2 = 2 √ 2 Equation of circle is ( x − 2 ) 2 + ( y − 3 ) 2 = 8

Equation of circle touching the line |x-2|+|y-3|=4 will be

Discussion Forum : Circles

Question -

Equation of circle touching the line |x-2|+|y-3|=4 will be

Options:

A .

(x - 2)^{2} + (y - 3)^{2} = 12

B .

(x - 2)^{2} + (y - 3)^{2} = 4

C .

(x - 2)^{2} + (y - 3)^{2} = 10

D .

(x - 2)^{2} + (y - 3)^{2} = 8

Answer: Option D
:
D

Perpendicular distance from centre to tangent = radius

$r = \frac{| 2 + 3 - 9 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2}$ Equation of circle is $(x - 2)^{2} + (y - 3)^{2} = 8$

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