# Orthogonal Circles and Condition of Orthogonal Circles

Here you will learn what are orthogonal circles and condition of orthogonal circles.

Let’s begin –

## Orthogonal Circles

Let two circles are $$S_1$$ = $${x}^2 + {y}^2 + 2{g_1}x + 2{f_1}y + {c_1}$$ = 0 and $$S_2$$ = $${x}^2 + {y}^2 + 2{g_2}x + 2{f_2}y + {c_2}$$ = 0.Then

Angle of intersection of two circles is cos$$\theta$$ = |$$2{g_1}{g_2} + 2{f_1}{f_2} – {c_1} – {c_2}\over {2\sqrt{{g_1}^2 + {f_1}^2 -c_1}}{\sqrt{{g_1}^2 + {f_1}^2 -c_1}}$$|

If the angle of intersection of the two circles is a right angle then such circles are called ‘Orthogonal circles’

i.e. $$\theta$$ = 90 $$\implies$$   $$cos\theta$$ = 0

$$\implies$$  |$$2{g_1}{g_2} + 2{f_1}{f_2} – {c_1} – {c_2}\over {2\sqrt{{g_1}^2 + {f_1}^2 -c_1}}{\sqrt{{g_1}^2 + {f_1}^2 -c_1}}$$| = 0

#### Orthogonal Circles Condition

$$2g_1g_2 + 2f_1f_2 = c_1 + c_2$$

Example : Show that the circles $$x^2 + y^2 + 4x + 6y + 3$$ = 0 and $$2x^2 + 2y^2 + 6x + 4y + 18 = 0$$ intersect orthogonally.

Solution : We have given two circles,

$$x^2 + y^2 + 4x + 6y + 3$$ = 0 and $$2x^2 + 2y^2 + 6x + 4y + 18 = 0$$

Write second circle in standard form by dividing by 2,

$$x^2 + y^2 + 3x + 2y + 9$$ = 0

Now, comparing these two equations with above equations $$S_1$$ and $$S_2$$,

From first equation,

$$g_1$$ = 2, $$f_1$$ = 3 and $$c_1$$ = 3

From second equation,

$$g_2$$ = $$3\over 2$$, $$f_2$$ = 1 and $$c_2$$ = 9

Condition to prove orthogonal,

$$2g_1g_2 + 2f_1f_2 = c_1 + c_2$$

2(2)($$3\over 2$$) + 2(3)(1) = 3 + 9

6 + 6 = 3 + 9

12 = 12

L.H.S = R.H.S

Hence, both the circle intersect orthogonally.