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Question

(B) The given circle is $x^{2} + y^{2} - 8x - 8y - 4 = 0$.The center of this circle is $(4, 4)$ and its radius $r = \sqrt{4^{2} + 4^{2} - (-4)} = \sqr

Vedclass · Accepted Answer

a parabola

Vedclass · Answer

(B) The given circle is $x^{2} + y^{2} - 8x - 8y - 4 = 0$.The center of this circle is $(4, 4)$ and its radius $r = \sqrt{4^{2} + 4^{2} - (-4)} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.Let the center of the required circle be $(h, k)$ and its radius be $R$. Since the circle touches the $x$-axis,its radius $R = |k|$. Assuming the circle is above the $x$-axis,$R = k$.Since the circle touches the given circle externally,the distance between their centers is equal to the sum of their radii:$\sqrt{(h - 4)^{2} + (k - 4)^{2}} = 6 + k$.Squaring both sides:$(h - 4)^{2} + (k - 4)^{2} = (6 + k)^{2}$.Expanding the terms:$h^{2} - 8h + 16 + k^{2} - 8k + 16 = 36 + k^{2} + 12k$.Simplifying the equation:$h^{2} - 8h - 20k - 4 = 0$.Replacing $(h, k)$ with $(x, y)$,we get the locus:$x^{2} - 8x - 20y - 4 = 0$.This is the equation of a parabola.

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

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