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

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Byju's Answer

Standard XII

Mathematics

Slope Form of Tangent

Equation of c...

Question

$E q u a t i o n o f c i r c l e t o u c h i n g t h e l i n e | x - 2 | + | y - 3 | = 4 w i l l b e$

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Solution

The correct option is D

$P e r p e n d i c u l a r d i s t a n c e f r o m c e n t r e t o t a n g e n t = r a d i u s$

$r = \frac{| 2 + 3 - 9 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2} E q u a t i o n o f c i r c l e i s (x - 2)^{2} + (y - 3)^{2} = 8$

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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√2Equationofcircleis(x−2)2+(y−3)2=8

$E q u a t i o n o f c i r c l e t o u c h i n g t h e l i n e | x - 2 | + | y - 3 | = 4 w i l l b e$

The correct option is D

$P e r p e n d i c u l a r d i s t a n c e f r o m c e n t r e t o t a n g e n t = r a d i u s$

$r = \frac{| 2 + 3 - 9 |}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4 \sqrt{2}}{2} = 2 \sqrt{2} E q u a t i o n o f c i r c l e i s (x - 2)^{2} + (y - 3)^{2} = 8$