Question

A circle is centered at (−4, −6) and has a radius of 3. Which of the following is the equation for this circle? (6 points)

Answer 1

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-4}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{3}{ r} \\\\\\\ [x-(-4)]^2+[y-(-6)]^2=3^2\implies (x+4)^2+(y+6)^2=9$

Answer 2

Answer:

The equation of circle is $(x+4)^2+(y+6)^2=9$.

Step-by-step explanation:

Given information:

Center of circle = (-4,-6)

Radius of circle = 3 units

The standard equation of a circle is

$(x-h)^2+(y-k)^2=r^2$            .... (1)

where, (h,k) is center of the circle and r is radius.

Substitute h=-4, k=-6 and r=3 in equation (1), to find the standard form of the given circle.

$(x-(-4))^2+(y-(-6))^2=(3)^2$

$(x+4)^2+(y+6)^2=9$

Therefore the equation of circle is $(x+4)^2+(y+6)^2=9$.