How do you write the standard form of the equation of the circle that is tangent to x=-2 and has its center at (2, -4)?

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Answer 1 · 2016-06-14T12:53:21

Equation of circle is $x^2+y^2-4x+8y+4=0$

Distance from a point $(m,n)$ to a line $ax+by+c=0$ is given by

$(am+bn+c)/sqrt(a^2+b^2)$

Hence, length of perpendicular from $(2,-4)$ to $x=-2$ or $x+2=0$ is $(2+2)/1=4$

Hence $4$ is radius and as center is $(2,-4)$ , equation of circle will be

$(x-2)^2+(y+4)^2=16$ or

$x^2-4x+4+y^2+8y+16=16$

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