The circle below is centered at the point (3,-4) and has a radius of length 3.
What is its equation?
-5
D. (x-3)² + (y-4)² = 9
A. (x-4)² + (y+ 3)² =
32
B. (x+4)² + (v- 3)² =
32
C. (x-3)² + (y + 4)² = 9



Answer :

The equation of the circle is given by (C) (x-3)² + (y + 4)² = 9

All points in a plane that are at a specific distance from a specific point, the center, form a circle. In other words, it is the curve that a moving point in a plane draws to keep its distance from a specific point constant. The radius of a circle is the separation between any point on the circle and its center.

The equation of the circle in the general form is given by the

(x-a)²+(y-b)²=r²

Where (a,b) is the co-ordinates of the center of the circle and the radius is r.

Now the center of the circle is given at (3,-4) and the radius is r.

We know that the general equation of a circle is  [tex]{\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}}[/tex]

Substituting the values in the equation we get:

(x-3)²+(y-(-4))²=(3)²

or, (x-3)²+(y+4)²=9

Hence the equation of the circle is (x-3)² + (y + 4)² = 9.

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