# Coordinate Geometry Circle

A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.

The fixed point is called the centre of the circle & the distance from the centre to a point on the circle is called the radius of the circle. Given $$C(h,k)$$ be the centre & r be the radius of circle. Let $$P(x,y)$$ be any point on the circle. $$|CP|=r$$. By distance formula:

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

Therefore, any point on the circle satisfies

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

Problem: Find the equation of the circle with centre (2,3) & radius 5?

Solution: Equation of the circle is

$${ (x-2) }^{ 2 }+{ (y-3) }^{ 2 }={ 5 }^{ 2 }\\ { x }^{ 2 }-4x+4+{ y }^{ 2 }+9-6y=25\\ { x }^{ 2 }+{ y }^{ 2 }-4x-6y-12=0$$

Important formula:

Let generalized equation of a circle be

$${ x }^{ 2 }+{ y }^{ 2 }+2gx+2fy+c=0$$

Centre of circle is at $$(-g,-f)$$

$$\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c }$$

Problem: Find the centre & the radius of the circle

$${ x }^{ 2 }+{ y }^{ 2 }-12x+10y+52=0$$

Solution: Let’s compare this equation with generalized equation:

$${ x }^{ 2 }+{ y }^{ 2 }-12x+10y+52=0$$

2g = -12, c = 52, 2f = 10

g = -6, f = 5, c = 52

Centre (-g,-f) = (6,-5)

Radius = $$\sqrt { { g }^{ 2 }+{ f }^{ 2 }-c } =\sqrt { 36+25-52 }$$