The centres of those circles which touch the circle, $x^{2} + y^{2} - 8 x - 8 y - 4 = 0$ , externally and also touch the x - axis, lie on

an ellipse which is not a circle

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a hyperbola

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a parabola

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a circle

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Solution

The correct option is C
a parabola

Radius $= \sqrt{16 + 16 + 4} = 6$

$(6 + k)^{2} = (h - 4)^{2} + (k - 4)^{2} R e p l a c e h \to x, k \to y (y + 6)^{2} - (y - 4)^{2} = x^{2} - 8 x + 16 (2 y + 2) (10) = x^{2} - 8 x + 16 20 y + 20 = x^{2} - 8 x + 16 x^{2} - 8 x - 20 y - 4 = 0$

Centre lies on parabola

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