What is the equation of the circle with a center at $(- 4, 6)$ $(-4 ,6 )$ and a radius of $3$ $3$ ?

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Answer 1 · 2016-07-25T17:59:38

$x^{2} + y^{2} + 8 x - 12 y + 43 = 0$ $x^2+y^2+8x-12y+43=0$

Eqn. of a circle with centre $(h, k) and r a d i u s = r$ $(h,k) and radius=r$ is given by,

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

In our case, the eqn. is, ${(x + 4)}^{2} + {(y - 6)}^{2} = 3^{2}$ $(x+4)^2+(y-6)^2=3^2$ , i.e.,

$x^{2} + y^{2} + 8 x - 12 y + 43 = 0$ $x^2+y^2+8x-12y+43=0$

What is the equation of the circle with a center at (-4 ,6 )(−4,6)(-4 ,6 ) and a radius of 3 33 ?