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Question

The locus of the centre of a circle, which touches externally the circle $$x^{2} + y^{2} - 6x - 6y + 14 = 0$$ and also touches the $$y$$-axis, is given by the equation


A
x26x10y+14=0
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B
x210x6y+14=0
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C
y26x10y+14=0
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D
y210x6y+14=0
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Solution

The correct option is D $$y^{2} - 10x - 6y + 14 = 0$$
Given the equation of circle is $$x^2+y^2-6x-6y+14=0$$

We know $$c = f^{2}$$

center$$ (3,3)$$.....From equation

and $$(g - 3)^{2} + (f - 3)^{2} = [\sqrt {9 + 9 - 14} + \sqrt {g^{2} + f^{2} - c^{2}}]^{2}$$

$$\Rightarrow g^{2} + f^{2} - 6g - 6f + 18 = (2 + g)^{2}; \text{as}\ c = f^{2}$$

$$\Rightarrow f^{2} - 10g - 6f + 14 = 0$$

Therefore, locus of centre $$(g, f)$$ is $$y^{2} - 10x - 6y + 14 = 0$$.

Mathematics

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