How do you write an equation of a circle with center at (1,-4), #r=sqrt17#?

1 Answer
Shwetank Mauria
Dec 2, 2016

The equation of desired circle is #x^2+y^2-2x+8y=0#

Explanation:

Circle is the locus of a point, which moves so that its distance from a given point called center is always same. This distance is called radius.

Usually we use a compass to draw a circle, but in a Cartesian plane, when the point is given as #(1,-4)# and radius is #sqrt17#, this means tracing a point #(x,y)#, whose distance from #(1,-4)# is always #sqrt17#.

As distance between two points #(x_1,y_1)# and #(x_2,y_2)# in a Cartesian plane is given by #sqrt((x_2-x_1)^2+(y_2-y_1)^2)#, as the distance between #(x,y)# and #(1,-4)# is #sqrt17#, we have

#sqrt((x-1)^2+(y-(-4))^2)=sqrt17# and squaring both sides we get

#(x-1)^2+(y+4))^2=17#

or #(x^2-2x+1)+(y^2+8y+16)=17#

or #x^2+y^2-2x+8y=0#, which is the equation of desired circle.
graph{x^2+y^2-2x+8y=0 [-9.04, 10.96, -8.44, 1.56]}

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