Two circles are said to be orthogonal circles if the tangent at their point of intersection are at right angles.

If two circles are cut orthogonally then it must satisfy the following condition.

**2 g₁ g₂ + 2 f₁ f₂ = c₁ + c₂**

Now let us see the example of orthogonal-circles

**Example 1 :**

Show that the circles x² + y² - 8x + 6y - 23 = 0 and x² + y² - 2x - 5y + 16 = 0 are orthogonal.

**Solution:**

First let us find the values of g₁, g₂,f₁ and f₂ from the equations by comparing the given equation with the general form of the circle

**x² + y² + 2gx + 2fy + c = 0**

x² + y² - 8x + 6y - 23 = 0

2g₁ = -8 2f₁ = 6 c₁ = -23

g₁ = -4 f₁ = 3

x² + y² - 2x - 5y + 16 = 0

2g₂ = -2 2f₂ = -5 c₂ = 16

g₂ = -1 f₂ = -5/2

Condition for orthogonal is

**2 g₁ g₂ + 2 f₁ f₂ = c₁ + c₂ **

2 (-4) (-1) + 2 (3) (-5/2) = -23 + 16

8 - 15 = -7

- 7 = -7

So the given condition is satisfied.So the given circles are orthogonal.

**Example 2 :**

Show that the circles x² + y² - 8x + 6y + 21 = 0 and x² + y² - 2y - 15 = 0 are orthogonal.

**Solution:**

First let us find the values of g₁, g₂,f₁ and f₂ from the equations by comparing the given equation with the general form of the circle

**x² + y² + 2gx + 2fy + c = 0**

x² + y² - 8x - 6y + 21 = 0

2g₁ = -8 2f₁ = -6 c₁ = 21

g₁ = -4 f₁ = -3

x² + y² - 2y - 15 = 0

2g₂ = 0 2f₂ = -2 c₂ = -15

g₂ = 0 f₂ = -1

Condition for orthogonal is

**2 g₁ g₂ + 2 f₁ f₂ = c₁ + c₂ **

2 (-4) (0) + 2 (-3) (-1) = 21 - 15

0 + 6 = 6

6 = 6

So the given condition is satisfied.So the given circles are orthogonal.

orthogonal circles

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