Orthogonal Circles
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Direct answer
Cutting orthogonally is a precise circle condition: two circles meet at right angles exactly when d² = r₁² + r₂², where d is the distance between centres — the Pythagorean relation applied to the two radii drawn to an intersection point, since each radius is perpendicular to its circle's tangent and orthogonality makes the tangents perpendicular. In general form, for S = x² + y² + 2gx + 2fy + c = 0 and S' = x² + y² + 2g'x + 2f'y + c' = 0, the condition is 2gg' + 2ff' = c + c'. Orthogonality is symmetric, and every coaxal family of circles possesses a partner coaxal family orthogonal to each of its members — the structure through which inversive geometry reaches JEE.
What you must remember
- Centre-radius form: d² = r₁² + r₂² with d the centre distance; if d = |r₁ − r₂| the circles touch internally, d = r₁ + r₂ externally, |r₁ − r₂| < d < r₁ + r₂ gives two intersections.
- General-form condition: 2gg' + 2ff' = c + c'; a derivation-free memory hook — "half the cross terms equals the constants' half".
- Radius from general form: r = √(g² + f² − c) must be real for the circle to exist; an "orthogonality" question sometimes first requires checking both circles are real.
- Orthogonal coaxal systems: the coaxal family x² + y² + 2λx + c = 0 (fixed c, common radical axis x = 0) is cut orthogonally by every member of the partner family x² + y² + 2μy − c = 0 — verify with the condition: 2(λ)(0) + 2(0)(μ) = c + (−c) = 0.
- Line as degenerate case: a line through the centre of a circle cuts it orthogonally — the "circle of infinite radius" reading.
- Reflection property: inversion in an orthogonal circle maps the given circle to itself — the reason orthogonal circles anchor inversion problems.
Testing orthogonality numerically
Test whether x² + y² = 16 and x² + y² − 6x − 8y + 16 = 0 are orthogonal. Circle 1 has centre (0, 0), r₁ = 4; circle 2 has centre (3, 4) and r₂ = √(9 + 16 − 16) = 3. The centre distance d = 5, and d² = 25 = 16 + 9 = r₁² + r₂² — orthogonal. The general-form check agrees: 2(0)(−3) + 2(0)(−4) = 0 = c + c' = −16 + 16. Two checks, one verdict — run whichever suits the data.
Now the reverse engineering that exams prefer: find the circles through the origin orthogonal to x² + y² − 4x − 6y + 8 = 0. A circle through the origin has c = 0: x² + y² + 2gx + 2fy = 0. Orthogonality demands 2g(−2) + 2f(−3) = 8 + 0, i.e. −4g − 6f = 8, one linear condition on two unknowns — a one-parameter family of answers unless a second point is specified. The recognition that orthogonality is a single linear condition on the circle's coefficients is the powerful fact: it never requires solving quadratics.
Two forms, one condition
Main tests the condition directly; Advanced embeds orthogonality in families — all circles through (2, 3) orthogonal to two given circles is two linear conditions, unique circle, no quadratic ever — and in coaxal systems, where each pencil has a conjugate pencil sharing its limiting points. The classic trap is sign error in reading g, f: the coefficient of x is 2g, so 2g = −6 means g = −3; plugging "g = −6" produces a confident wrong answer the options anticipate. The second trap is forgetting that both radii must be real before any Pythagorean talk — an imaginary circle satisfies formal conditions vacuously. The signature picture: at each intersection, the two radii and two tangents form a square-cornered cross.
Frequently asked questions
What is the condition for two circles to cut orthogonally?
d² = r₁² + r₂² in centres-and-radii form, or 2gg' + 2ff' = c + c' in general form.
Why does d² = r₁² + r₂² express perpendicular tangents?
At an intersection point, each radius is perpendicular to its own tangent; if the two radii (meeting at the point, tails at the centres distance d apart) form a right triangle with hypotenuse d, the tangents are perpendicular too.
Is x² + y² − 6x − 8y + 16 = 0 orthogonal to x² + y² = 16?
Yes: centres (3, 4) and (0, 0) are 5 apart, and 5² = 3² + 4² = r₂² + r₁² with radii 3 and 4.
How many circles pass through a fixed point and are orthogonal to a given circle?
Infinitely many — orthogonality imposes one linear condition on the circle's coefficients, so with one point fixed, a one-parameter family remains.
What is the link between orthogonal circles and coaxal systems?
Each coaxal family has an orthogonal partner family sharing the same limiting points — the structure behind radical-axis and inversion questions.