Orthogonal intersection of two circles:

The 2 circles are said to intersect orthogonally if angle between the tangents at their point of intersection is equal to 90⁰.

The condition for 2 circles S₁=O and S₂=O to cut each other orthogonally is given by

Note:  It the 2 circles are intersecting orthogonally the tangent to one circle at the point of intersection passes through the centre of the other circle.

Case V:   It the distance between centres is less than the difference of their radii that is |O₁ O₂| < |r₁-r₂|, in this case one circle will lie inside the other circle. Therefore there will be no common tangent.

Illustration: If the circles x² + y² + 2ax + 2by = 0 and x² + y² + 2bx + 2cy = 0 touch each other, then show that b²=ac.

Solution : Both the circles are touching each other. Thus distance between the centres should be equal to sum of the radii. Centre are (-a, -b) and (-b, -c) and their radii are  respectively

                                                squaring both the sides

                                                => b⁴ - 2acb² + a²c² = 0       => (b² - ac)² = 0 => b² = ac

Solution: Clearly both the circles are passing through origin. Thus they will be ouching each other at (0, 0). Therefore at (0, 0) they will be having a common tangent.

                                                => 0.x + 0.y + a(x + 0) + b(y + 0) = 0 => ax + by = 0          ...(1)

                                                =>  0.x + 0.y + b(x + 0) + c(y + 0) = 0 => bx + cy = 0         ...(2)

                                                (1) and (2) identical

                                                a/b = b/c => b² = ac

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