Equation of circle passing through origin, (a, 0), (0, b):
Ax² + y² = a + b
BOther
Cx² + y² - ax - by = 0 (using general second-degree fits)
Dx² + y² = ab
Answer & Solution
Correct answer: C. x² + y² - ax - by = 0 (using general second-degree fits)
General circle x² + y² + 2gx + 2fy + c = 0. Through (0,0): c = 0. Through (a,0): a² + 2ag = 0 → g = -a/2. Through (0,b): b² + 2bf = 0 → f = -b/2. So x² + y² - ax - by = 0.
Related questions
The Cassegrain telescope design uses the reflective property of a:A satellite dish uses a parabolic reflector because:By Kepler's First Law, every planet orbits the Sun in:For a general 2nd-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, the curve is a PARABOThe HYPERBOLA x²/a² − y²/b² = 1 has asymptotes given by:In the standard ellipse x²/a² + y²/b² = 1 (with a > b), the relationship between a, b, c (The standard form of a circle with centre (3, -2) and radius 5 is:The eccentricity of a CIRCLE is: