Find the equation of circle which is concentric to $x^2+y^2-8x+4=0$ and touches line $x+2y+6=0$ ?

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Answer 1 · 2017-04-10T05:40:53

$x^2+y^2-8x-4=0$

The center of circle $x^2+y^2-8x+4=0$ is $(4,0)$

As the desired circle is concentric to this circle, its center too is $(4,0)$

and radius is the length of perpendicular from $(4,0)$ to $x+2y+6=0$

i.e. $|(4+2xx0+6)/(sqrt(1^2+2^2))|=10/sqrt5$

Hence equation of circle is

$(x-4)^2+y^2=100/5=20$

or $x^2-8x+16+y^2-20=0$

or $x^2+y^2-8x-4=0$

graph{(x+2y+6)(x^2+y^2-8x-4)(x^2+y^2-8x+4)=0 [-10, 10, -5, 5]}

Find the equation of circle which is concentric to x^2+y^2-8x+4=0x^2+y^2-8x+4=0 and touches line x+2y+6=0x+2y+6=0?