If a circle passes through the point (a, b) and cuts the circle x^2 + y^2 = p^2 orthogonally, then the equation of the locus of its centre is: - Mathematics MCQ AIEEE 2026 | ExamTest.live

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If a circle passes through the point (a, b) and cuts the circle x^{2} + y^{2} = p^{2} orthogonally, then the equation of the locus of its centre is:

Solution & Step-by-step Explanation

Let the required circle be x^{2} + y^{2} - 2 h x - 2 k y + c = 0 .It cuts x^{2} + y^{2} - p^{2} = 0 orthogonally.Condition for orthogonality 2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2} : 2 (- h) (0) + 2 (- k) (0) = c - p^{2} ⟹ c = p^{2} .The circle equation is x^{2} + y^{2} - 2 h x - 2 k y + p^{2} = 0 .It passes through (a, b) : a^{2} + b^{2} - 2 ha - 2 kb + p^{2} = 0 Replacing (h, k) with (x, y) for the locus: a^{2} + b^{2} - 2 x a - 2 y b + p^{2} = 0 2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0 .

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If a circle passes through the point (a, b) and cuts the circle x^{2} + y^{2} = p^{2} orthogonally, then the equation of the locus of its centre is:

x^{2} + y^{2} - 3 a x - 4 b y + (a^{2} + b^{2} - p^{2}) = 0

2 a x + 2 b y - (a^{2} - b^{2} + p^{2}) = 0

x^{2} + y^{2} - 2 a x - 3 b y + (a^{2} - b^{2} - p^{2}) = 0

2 a x + 2 b y - (a^{2} + b^{2} + p^{2}) = 0

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