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.

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