Equation of circle touching the line -x-2-y-3-4 will beA- x-22-y-32-12B- x-22-y-32-10C- x-22-y-32-4D- x-22-y-32-8

The correct option is 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 ...

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Byju's Answer

Standard IX

Mathematics

Equation of Circle with (h,k) as Center

Equation of c...

Question

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

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Solution

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

The correct option is D Perpendicular distance from centre to tangent = radius r=|2+3−9|√2=4√2=4√22=2√2 Equation of circle is (x−2)2+(y−3)2=8

The correct option is 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 $(x - 2)^{2} + (y - 3)^{2} = 8$