# General Form of a Circle

The equation x² +y²** +2 gx +2 fy +c = 0 represents a circle iff g² +f² -c > 0.**

**Its center is (-g, -f) and radius = [g² +f² -c].**

**Concentric circles.** Circles having same center are called **concentric circles.**

**Equal circles** Circles having equal radius are called **equal circles.**

## Illustrative Examples

### Example

One end of a diameter of the circle x² +y² -6 x +5 y -7 = 0 is (-1, 3). Find the co-ordinates of the other end.

### Solution

The given equation is

x² + y² -6 x +5 y -7 = 0

It is easy to see that this represents a circle with center C (3, - 5/2)

Let A(-1,3) and B(,) be the ends of
the diameter.

Since C is mid-point of [AB], we get

( - 1)/2 = 3 and
( +3) / 2 = - 5/2

= 7 and = -8

Hence the other end of the diameter is (7, -8).

### Example

Show that the points (7, 5), (6, -2), (-1, -1) and (0, 6) are concyclic. Also find the radius and the center of the circle on which they lie.

### Solution

Let us find the equation of the circle passing through the points(7, 5),
(6, - 2) and (-1, -1).

Let the equation of this circle be

x² + y² + 2gx + 2fy + c = 0
...(i)

As the points (7, 5), (6, -2) and (-1, -1) lie on it, we get

49 +25 +14 g +10 f +c => 14 g +10 f +c +74 = 0 ...(ii)

36 +4 +12 g -4 f +c = 0 => 12 g -4 f + c +40 = 0 ...(iii)

1 -2 g -2 f +c = 0 => 2 g +2 f -c -2 = 0 ...(iv)

Adding (ii) and (iv), we get

16 g +12 f +72 = 0 => 4 g +3 f +18 = 0 ...(v)

Adding (iii) and (iv), we get

14 g -2 f +38 = 0 => 7 g - f +19 = 0 ...(vi)

Solving (v) and (vi) simultaneously, we get g = -3, f = -2. ...(vi)

From (ii), we get c = -14 (-3) -10 (-2) -74 = -12.

Substituting these values of g, f and c in (i), we get

x² +y² -6 x -4 y -12 = 0 ...(vii)

The fourth point (0, 6) will lie on (vii) if 0 +36 -0 -24 -12 = 0 i.e. if 0 = 0, which is true.

Hence the given points are concyclic.

Also, (vii) is the equation of the circle on which these points lie.

Its center is (3, 2) and radius = [9 +4 -(-12)] = 5.

## Exercise

- Which of the following equations represent a circle? If so, determine its center and radius:

(i) x² + y² +4 x -4 y -1 = 0

(ii) 2 x² +2 y² = 3 x -5 y +7

(iii) x² +y² +4 x +2 y +14 = 0

(iv) 2 x² +2 y² = 5 x +7 y + 3

(v) (x +3)² +(y -2)² = 0

(vi) x² + y² - a x - b y = 0 - Find the value of p so that x² + y² +8 x; +10 y +p = 0 is the equation of a circle of radius 7 units.
- The radius of the circle x² +y² -2 x +3 y +k = 0 is 2. Find the value of k. Find also the equation of the diameter of the circle which passes through the point.
- (i) Find the equation of the circle the end points of whose one diameter
are the centers of the circles x² +y² +6 x -14 y +5 = 0 and x² + y² -4 x +10 y +7 = 0.

(ii) One end of a diameter of the circle x² + y² -3 x +5 y -4 = 0 is (2, 1), find the co-ordinates of the other end. - Find the equation of the circle concentric with the circle x² +y² -8 x +6 y -5 = 0 and passing through the point (-2, -7).
- Find the equation of the circle which passes through the center of the circle x² +y² -4 x -8 y -41 = 0 and is concentric with the circle x² +y² -2 y +1 = 0.
- Find the equation of the circle concentric with the circle 2 x² +2 y² +8 x +10 y -35 = 0 and with area 16 square units.
- Find the equation of the circle which is concentric with the circle x² + y² -4 x +6 y -3 = 0 and of double its (i) circumference (ii) area.
- Prove that the centres of three circles x² + y² -4 x -6 y -14 = 0, x² + y² +2 x +4 y -5 = 0 and x² + y² -10 x -16 y +7 = 0 are collinear.
- Prove that the radii of the circles x² + y² = 1, x² + y² -2 x -6 y -6 = 0 and x² + y² -4 x -12 y -9 = 0 are in A.P.
- Find the equation of the circle which has its center on the line y = 2 and which passes through the points (2, 0) and (4, 0).
- Find the equation of the circle which passes through the points (1, - 2), (4, -3) and has its center on the line 3 x +4 y +10 = 0.
- Find the equation of the circle passing through the three points

(i) (0, 0), (0, 1) and (2, 3)

(ii) (0, 2), (3, 0) and (3, 2);

Also find its center and radius.

## Answers

**1.**(i) circle; (- 2 , 2), 3 (ii) circle; (3/4, 5/4), (310)/4 (iii) empty set

(iv) circle; (5/4, 7/4), (v) point circle; (-3, 2), zero

(vi) circle;(a/2, b/2)

**2.**-8

**3.**-3; 2 x -2 y = 5

**4.**(i) x² + y² +x -2 y -41 = 0 (ii) (1, -6)

**5.**x² +y² -8 x +6 y -27 = 0

**6.**x² + y² -2 y -12 = 0

**7.**4 x² +4 y² +16 x +20 y -23 = 0

**8.**(i) x² + y² -4 x +6 y -51 = 0 (ii) x² + y² -4 x +6y -19 = 0

**11.**x² +y² -6 x -4 y +8 = 0

**12.**x² +y² -4 x +8 y +15 = 0

**13.**(i) x² +y² -5 x - y = 0 -y = 0 ; (5/2, 1/2), 26 /2

(ii) x² +y² -3 x -2 y = 0 ; (3/2, 1), 13 /2