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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