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Identify the centre and radius from a circle equation

A circle in standard form, (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2, is interpreted as having centre (h,k)(h,k) and radius rr, with rr equal to the nonnegative square root of the constant on the right. Careful attention is given to the opposite signs inside the squared binomials, so (x+3)2(x+3)^2 places the centre at x=−3x=-3; this scope concerns real circles in the coordinate plane, not general conic, parametric, or three-dimensional forms.

Detailed Explanation: Identify the centre and radius from a circle equation

A circle in standard form is written as

(x−h)2+(y−k)2=r2,(x-h)^2+(y-k)^2=r^2,

where:

  • (h,k)(h,k) is the centre
  • rr is the radius

To identify the centre, look carefully at the signs inside the brackets. The signs are opposite to the coordinates of the centre.

Example

Find the centre and radius of

(x+3)2+(y−4)2=25.(x+3)^2+(y-4)^2=25.

Step 1: Identify the centre.

Compare the equation with

(x−h)2+(y−k)2=r2.(x-h)^2+(y-k)^2=r^2.

For the xx-part,

(x+3)2=(x−(−3))2,(x+3)^2=(x-(-3))^2,

so h=−3h=-3.

For the yy-part,

(y−4)2(y-4)^2

shows that k=4k=4.

Therefore, the centre is

(−3,4).\boxed{(-3,4)}.

Step 2: Identify the radius.

The number on the right is r2r^2:

r2=25.r^2=25.

The radius is the nonnegative square root:

r=25=5.r=\sqrt{25}=5.

Therefore, the radius is

5.\boxed{5}.

So the circle has centre (−3,4)\boxed{(-3,4)} and radius 5\boxed{5}. Remember that a plus sign inside a bracket gives a negative coordinate: (x+3)2(x+3)^2 means x=−3x=-3.

Learn by doing: Identify the centre and radius from a circle equation

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