# Orthogonal Circles

### Angle of intersection of two circles

An angle between the two tangents to the two circles at a point of
intersection is called an **angle of intersection between two circles.**

### Orthogonal circles

Two circles are said to **cut orthogonally** iff angle of intersection
of these circles at a point of intersection is a right angle i.e. iff the
tangents to these circles at a common point are perpendicular to each other.

If r, r' are radii of circles S and S' respectively and d is the distance
between their centers and is an angle of intersection of
these circles, then

cos =(r² +r'² -d²)/(2 r r')

If the circles S and S' cut orthogonally, then = 90°,cos
= cos 90° = 0

=> (r² +r'² -d²)/(2r r') = 0

=> r² +r'² = d²

Two circles S = x² +y² +2gx +2 f y +c = 0 and

S' = x² +y² +2 g' x +2 f'y +c' = 0 cut orthogonally if 2 (g g' +f f') = c +c'.

## Illustrative Examples

### Example

Find the equation of the circle which intersects the circles x² +y² +2 x -2 y +1 = 0 and x² +y² +4 x -4 y +3 = 0 orthogonally and whose center lies on the line 3 x -y -2 = 0.

### Solution

Let the equation of the required circle be

x² +y² +2 g x +2 f y + c = 0 ...(i)

Since this circle cuts the circles x² +y² +2 x -2 y +1 = 0 and x² + y²
+4 x -4 y +3 = 0 orthogonally, we get

2 (g. 1 + f. (-1)) = c +1

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

and 2 (g.2 + f.(-2)) = c +3

=> 4 g -4 f -c -3 = 0
...(iii)

As center of (i) i.e. (-g, -f) lies on 3 x -y -2 = 0, we get

-3 g + f -2 = 0

=> 3 g - f +2 = 0
...(iv)

Subtracting (ii) from (iii), we get

2 g -2 f -2 = 0

=> g -f -1 = 0
...(v)

Solving (iv) and (v) simultaneously, we get g = -3/2 and f = -5/2

From (ii), we get c = 2 g -2 f -1 = -3 +5 -1 = 1

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

x² +y² -3 x -5 y +1 = 0, which is the equation of the required circle.

## Exercise

- Determine the angle of intersection of the two circles (x -3)² +(y -1)² = 8 and (x -2)² +(y +2)² = 2.
- Find the angle at which the circles x² + y² = 16 and x² +y² -2 x -4 y = 0 intersect each other.
- Show that the circles x² + y² -4 x -6 y +4 = 0 and x² +y² -10 x -14 y +58 = 0 cut orthogonally.
- Show that the circles x² +y² -2 a x +2 b y + c = 0 and x² + y² +2 bx +2 a y -c = 0, a² + b² > | c |, cut orthogonally.
- For what value of do the circles x² + y² +5 x +3 y +7 = 0 and x² +y² -8 x +6 y + = 0 cut orthogonally?

## Answers

**1.**/2

**2.**The acute angle is given by cos = 2/ 5

**5.**-18