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  • Find the equation of a circle which touches both the axes and the line  <img alt="\mathrm{ 3 x-4 y+8=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%203%20x-4%20y&am

Find the equation of a circle which touches both the axes and the line  \mathrm{ 3 x-4 y+8=0}  and lies in the third quadrant.

Option: 1

\mathrm{x^2+y^2+4 x-4 y-4=0 }


Option: 2

\mathrm{x^2+y^2+4 x+4 y+4=0 }


Option: 3

\mathrm{x^2+y^2-4 x-4 y+4=0 }


Option: 4

\mathrm{x^2+y^2+4 x+4 y-4=0}


Answers (1)

best_answer

Let a be the radius of the circle, then (– a, – a) will be centre and perpendicular distance from the centre to the given line gives the radius of the circle

Since \mathrm{d=\left|\frac{-3 a+4(a)+8}{\sqrt{9+16}}\right|=\left|\frac{a+8}{5}\right|= \pm\left(\frac{a+8}{5}\right)}
\mathrm{\therefore a=\frac{a+8}{5} ~and ~a=\frac{-a-8}{5}}
\mathrm{\Rightarrow a=2 ~and ~a=-4 / 3}
But \mathrm{a \neq-4 / 3 \Rightarrow a=2}
Hence required equation of circle is

\mathrm{(x+2)^2+(y+2)^2=2^2 }
\mathrm{\Rightarrow }   \mathrm{x^2+y^2+4 x+4 y+4=0 } 
 

Posted by

avinash.dongre

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