Solve the problem.
-Write an equation of the circle that is tangent to both axes with radius $\sqrt { 3 }$ and center in Quadrant IV.
A) $( x + \sqrt { 3 } ) ^ { 2 } + ( y - \sqrt { 3 } ) ^ { 2 } = \sqrt { 3 }$
B) $( x - \sqrt { 3 } ) ^ { 2 } + ( y + \sqrt { 3 } ) ^ { 2 } = 3$
C) $( x - \sqrt { 3 } ) ^ { 2 } + ( y + \sqrt { 3 } ) ^ { 2 } = \sqrt { 3 }$
D) $( x + \sqrt { 3 } ) ^ { 2 } + ( y - \sqrt { 3 } ) ^ { 2 } = 3$

Question

Quizplus · Accepted Answer

The center of the circle in Quadrant IV with a radius of $\sqrt{3}$ would be at $(\sqrt{3}, -\sqrt{3})$, since it's tangent to both axes and in Quadrant IV. The general equation of a circle is $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius. Substituting $h = \sqrt{3}$, $k = -\sqrt{3}$, and $r = \sqrt{3}$, we get $(x - \sqrt{3})^2 + (y + \sqrt{3})^2 = (\sqrt{3})^2$, which simplifies to $(x - \sqrt{3})^2 + (y + \sqrt{3})^2 = 3$.

Solve the Problem 3\sqrt { 3 }3​ And Center in Quadrant IV

Solve the Problem $\sqrt { 3 }$ And Center in Quadrant IV