Circle 4 Question 9

9. The locus of the centre of a circle, which touches externally the circle $x^{2}+y^{2}-6 x-6 y+14=0$ and also touches the $Y$-axis, is given by the equation $(1993,1 M)$

(a) $x^{2}-6 x-10 y+14=0$

(b) $x^{2}-10 x-6 y+14=0$

(c) $y^{2}-6 x-10 y+14=0$

(d) $y^{2}-10 x-6 y+14=0$

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Answer:

Correct Answer: 9. (d)

Solution:

  1. Let $(h, k)$ be the centre of the circle which touches the circle $x^{2}+y^{2}-6 x-6 y+14=0$ and $Y$-axis.

The centre of given circle is $(3,3)$ and radius is $\sqrt{3^{2}+3^{2}-14}=\sqrt{9+9-14}=2$

Since, the circle touches $Y$-axis, the distance from its centre to $Y$-axis must be equal to its radius, therefore its radius is $h$. Again, the circles meet externally, therefore the distance between two centres $=$ sum of the radii of the two circles.

Hence,

$$ (h-3)^{2}+(k-3)^{2}=(2+h)^{2} $$

$$ h^{2}+9-6 h+k^{2}+9-6 k=4+h^{2}+4 h $$

i.e.

$$ k^{2}-10 h-6 k+14=0 $$

Thus, the locus of $(h, k)$ is

$$ y^{2}-10 x-6 y+14=0 $$



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